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Engineer Radio Electronic Communication study materials
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Engineer Radio Electronic Communication practice questions

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Question 1Radio wave propagation and antennas

A half-wave dipole is operated at 100 MHz. What is its approximate physical length in free space?

  1. 0.75 m
  2. 3.00 m
  3. 1.50 m Correct answer.
  4. 6.00 m

Explanation

Wavelength equals the speed of light divided by frequency, so 300/100 MHz gives 3 m. A half-wave dipole is about half a wavelength, i.e. roughly 1.5 m before any shortening factor is applied.

Question 2Radio wave propagation and antennas

Which propagation mode supports long-distance HF communication at night by refracting signals between the ionosphere and the ground?

  1. Line-of-sight propagation
  2. Tropospheric scatter
  3. Sky-wave propagation Correct answer.
  4. Ground-wave propagation

Explanation

Sky-wave propagation relies on refraction in the ionosphere, mainly the F layer at night, allowing beyond-horizon HF links via successive hops. Ground wave and line-of-sight are limited to shorter ranges, and tropospheric scatter uses the lower atmosphere.

Question 3Radio wave propagation and antennas

The characteristic impedance of free space is approximately:

  1. 120 ohms
  2. 75 ohms
  3. 50 ohms
  4. 377 ohms Correct answer.

Explanation

Free space has an intrinsic impedance of about 377 ohms, equal to the square root of permeability divided by permittivity. This value sets the relationship between electric and magnetic field strengths in a plane wave.

Question 4Radio wave propagation and antennas

A transmitting antenna has a gain of 10 dBi and the receiving antenna has a gain of 3 dBi. Ignoring losses and path loss, what is the total system gain in dB?

  1. 30 dB
  2. 7 dB
  3. 3.3 dB
  4. 13 dB Correct answer.

Explanation

In decibels, gains multiply by adding, so 10 dBi plus 3 dBi gives 13 dB of combined antenna gain. The dBi reference cancels when both antennas use the same isotropic reference.

Question 5Radio wave propagation and antennas

Which layer of the ionosphere is primarily responsible for long-distance HF sky-wave refraction during daytime?

  1. E layer
  2. F layer Correct answer.
  3. Troposphere
  4. D layer

Explanation

The F layer, especially the merged F2 region by day, has the highest electron density and supports the greatest refraction for long HF hops. The D layer mainly absorbs, and the troposphere is below the ionosphere.

Question 6Radio wave propagation and antennas

What is the primary effect of the ground plane on a quarter-wave vertical monopole?

  1. It converts the antenna into a loop radiator.
  2. It reduces the antenna's resonant frequency to zero.
  3. It provides the missing half of the dipole, shaping the radiation pattern. Correct answer.
  4. It eliminates all ground losses automatically.

Explanation

A quarter-wave monopole needs a conducting ground plane to act as the image of the missing dipole half. The ground plane thereby completes the radiating structure and produces a low-angle vertical pattern.

Question 7Radio wave propagation and antennas

Which phenomenon causes VHF signals to travel beyond the radio horizon by bending in the lower atmosphere?

  1. Faraday rotation
  2. Tropospheric ducting Correct answer.
  3. Doppler shift
  4. Ionospheric skip

Explanation

Tropospheric ducting occurs when temperature and humidity gradients create a refractive duct that traps VHF and UHF signals and carries them beyond the normal horizon. It is a lower-atmosphere effect, distinct from ionospheric skip.

Question 8Radio wave propagation and antennas

What polarization does a horizontal half-wave dipole produce in its broadside direction?

  1. Horizontal polarization Correct answer.
  2. Circular polarization
  3. Vertical polarization
  4. Elliptical polarization

Explanation

The electric field of a horizontal dipole is oriented along the wire, so the radiated wave is horizontally polarized broadside to the antenna. Circular and elliptical polarization require phase-quadrature or unequal orthogonal components.

Question 9Radio wave propagation and antennas

Two isotropic antennas are separated by 3 km at 300 MHz. What is the approximate free-space path loss?

  1. About 92 dB Correct answer.
  2. About 152 dB
  3. About 62 dB
  4. About 32 dB

Explanation

Using the standard formula, path loss in dB equals 32.45 plus 20 log10(f in MHz) plus 20 log10(d in km). That gives 32.45 + 49.5 + 9.5, approximately 92 dB.

Question 10Radio wave propagation and antennas

A transmission line has a 2:1 VSWR. What percentage of incident power is reflected?

  1. 11% Correct answer.
  2. 50%
  3. 0%
  4. 25%

Explanation

The reflection coefficient magnitude is (VSWR-1)/(VSWR+1), which is 1/3 for 2:1. Squaring gives about 0.111, so roughly 11 percent of incident power is reflected.

Question 11Radio wave propagation and antennas

A 50-ohm coaxial line feeds an antenna whose input impedance is 75 ohms resistive. What is the reflection coefficient magnitude?

  1. 0.75
  2. 0.20 Correct answer.
  3. 0.33
  4. 0.50

Explanation

The reflection coefficient is (ZL - Z0)/(ZL + Z0), which is 25/125 = 0.20. This corresponds to a VSWR of 1.5:1.

Question 12Radio wave propagation and antennas

Which statement best explains why the critical frequency of the ionosphere increases with solar activity?

  1. Solar activity changes only the Earth's magnetic field, not ionization.
  2. Higher solar activity increases ionization, raising the maximum usable frequency. Correct answer.
  3. Higher solar activity reduces the ionosphere's electron density.
  4. Critical frequency depends only on antenna height.

Explanation

Increased solar ultraviolet and X-ray flux raises ionospheric electron density, which increases the critical frequency and the maximum usable frequency for a given path. Antenna height and magnetic field changes are secondary or unrelated.

Question 13Radio wave propagation and antennas

A Yagi-Uda antenna achieves directivity mainly through which mechanism?

  1. A resistive termination absorbing backward radiation
  2. A resonant cavity enclosing the elements
  3. Parasitic elements acting as reflector and directors Correct answer.
  4. A ferrite core concentrating the magnetic field

Explanation

The Yagi uses a driven element plus a reflector and directors that couple parasitically to shape the radiation pattern toward the directors. It is not a cavity, ferrite, or terminated absorber design.

Question 14Radio wave propagation and antennas

What is the main advantage of a parabolic reflector antenna at microwave frequencies?

  1. It produces omnidirectional coverage.
  2. It eliminates the need for a feed antenna.
  3. It operates without any aperture.
  4. It provides high gain through a narrow, focused beam. Correct answer.

Explanation

A parabolic reflector collimates energy from the feed into a narrow beam, giving high gain and good directivity at microwave frequencies. It still requires a feed and has a finite aperture.

Question 15Radio wave propagation and antennas

A radio link at 2 GHz uses antennas with 20 dBi gain each. If the received power must be -80 dBm and the transmit power is 10 dBm, what is the maximum allowable path loss?

  1. 70 dB
  2. 130 dB Correct answer.
  3. 110 dB
  4. 150 dB

Explanation

The link budget gives received power equal to transmit power plus both antenna gains minus path loss. So -80 = 10 + 20 + 20 - L, giving L = 130 dB.

Question 16Radio wave propagation and antennas

Which effect describes the change in apparent frequency of a received signal caused by relative motion between transmitter and receiver?

  1. Doppler shift Correct answer.
  2. Fresnel reflection
  3. Group delay
  4. Multipath fading

Explanation

Relative motion between transmitter and receiver compresses or stretches the apparent wavelength, shifting the received frequency. This is the Doppler effect, important in mobile and satellite links.

Question 17Radio wave propagation and antennas

A dipole is shortened physically to resonate at a lower frequency. Which matching component is typically added to restore resonance?

  1. A shunt capacitor across the feed
  2. A balun transformer
  3. A loading coil in series Correct answer.
  4. A ferrite bead on the feed line

Explanation

Shortening a dipole raises its resonant frequency, so a series loading coil adds inductive reactance to cancel the capacitive reactance and restore resonance at the desired lower frequency. A balun and ferrite bead serve other purposes.

Question 18Radio wave propagation and antennas

What causes multipath fading on a line-of-sight microwave link?

  1. Absorption by the ionosphere
  2. Multiple reflected paths arriving with different phases Correct answer.
  3. Variations in the transmitter's power supply
  4. A change in the antenna's polarization only

Explanation

Multipath fading arises when the direct signal and reflected signals combine with varying phase, causing constructive or destructive interference. Power supply changes, pure polarization changes, and ionospheric absorption are not the cause.

Question 19Radio wave propagation and antennas

In a satellite link, which quantity is most directly improved by increasing the receiving antenna's effective aperture?

  1. The received signal power Correct answer.
  2. The orbital altitude
  3. The transmitter's EIRP
  4. The downlink frequency

Explanation

A larger effective aperture captures more of the incident power density, directly increasing received signal power for a given flux. It does not change the transmitter's EIRP, the orbit, or the assigned frequency.

Question 20Radio wave propagation and antennas

Which statement correctly describes the relationship between antenna directivity and gain?

  1. Gain is independent of efficiency.
  2. Gain equals directivity divided by efficiency.
  3. Gain and directivity are always identical.
  4. Gain equals directivity multiplied by efficiency. Correct answer.

Explanation

Antenna gain is directivity scaled by radiation efficiency, so gain equals directivity times efficiency. A lossless antenna has gain equal to directivity, but real antennas have efficiency below one.

Question 21Radio wave propagation and antennas

A 100 W transmitter feeds an antenna through a line with 3 dB loss. What is the power delivered to the antenna?

  1. 50 W Correct answer.
  2. 70 W
  3. 97 W
  4. 25 W

Explanation

A 3 dB loss halves the power, so 100 W becomes 50 W at the antenna. Decibel loss adds, and 3 dB corresponds to a factor of two.

Question 22Radio wave propagation and antennas

Which mechanism allows HF signals to propagate beyond the line of sight during daylight over very long distances?

  1. Ground wave along the sea surface
  2. Meteor-burst reflection
  3. Multiple hops between the F layer and the ground Correct answer.
  4. Tropospheric scattering at low altitudes

Explanation

Long daytime HF paths use successive refractions from the F layer and reflections from the ground, forming multihop sky-wave paths. Ground wave, tropospheric scatter, and meteor bursts serve shorter or intermittent roles.

Question 23Radio wave propagation and antennas

A circularly polarized receiving antenna is used to receive a linearly polarized signal. What is the resulting polarization mismatch loss?

  1. Infinite
  2. About 10 dB
  3. About 3 dB Correct answer.
  4. 0 dB

Explanation

A circularly polarized antenna receives only one linear component of a linearly polarized wave, so half the power is lost, about 3 dB. The mismatch is finite because one polarization component is fully captured.

Question 24Radio wave propagation and antennas

Why does the maximum usable frequency for a given ionospheric path increase as the path length increases up to a limit?

  1. Longer paths reduce free-space loss.
  2. The ionosphere's electron density rises with distance.
  3. Oblique incidence allows higher frequencies to be refracted at the same electron density. Correct answer.
  4. Longer paths require lower frequencies to avoid absorption.

Explanation

At oblique incidence, the secant law shows that a higher frequency can still be refracted by the same electron density because the ray enters at a shallower angle. This raises the maximum usable frequency until geometric limits are reached.

Question 25Radio wave propagation and antennas

A two-element array has elements fed with equal amplitude and 180 degrees phase difference, spaced a half wavelength apart. What is the resulting pattern?

  1. An omnidirectional pattern
  2. A broadside bidirectional pattern
  3. A bidirectional pattern with nulls broadside Correct answer.
  4. An end-fire unidirectional pattern

Explanation

With half-wavelength spacing and 180-degree phase difference, the fields add along the array axis and cancel broadside, producing a bidirectional pattern with nulls perpendicular to the array. This is the basis of the end-fire versus broadside distinction.

Question 26Radio wave propagation and antennas

A vertically polarized wave reflects from a smooth, dry, flat ground surface at a grazing angle. Which statement describes the reflected wave relative to the incident wave?

  1. The reflected wave is horizontally polarized because reflection rotates the plane of polarization.
  2. The reflected wave is vertically polarized and its electric field is reversed in phase relative to the incident wave at the surface. Correct answer.
  3. The reflected wave is vertically polarized and its phase is shifted by exactly 90 degrees.
  4. The reflected wave is circularly polarized because the ground introduces a quadrature component.

Explanation

For a vertically polarized wave incident on a ground plane at grazing incidence, the reflection coefficient approaches -1, so the reflected electric field is nearly equal in magnitude and reversed in phase at the boundary. The polarization remains vertical because the reflection does not rotate the plane of the electric field for a smooth planar surface. This phase reversal is why vertical antennas benefit from a conducting ground plane and why ground-reflected rays can cancel the direct ray at low elevation angles.

Question 27Radio wave propagation and antennas

A half-wave dipole is oriented along the z-axis. In which direction is its radiation pattern null (zero field)?

  1. In every direction, because a dipole is isotropic.
  2. In the equatorial plane perpendicular to the dipole axis.
  3. At 45 degrees from the dipole axis.
  4. Along the z-axis, the dipole axis. Correct answer.

Explanation

The current distribution on a dipole flows along its axis, so the radiated field is maximum broadside and falls to zero along the axis itself. The ideal half-wave dipole pattern has nulls at the two poles along the wire and a doughnut-shaped maximum in the plane perpendicular to it. This is why a vertical dipole radiates best toward the horizon rather than straight up.

Question 28Radio wave propagation and antennas

A ground-wave signal at 1 MHz must reach a receiver 150 km away over land of moderate conductivity. Which change most improves the received field strength at that range?

  1. Raising the transmit frequency to move the signal into the VHF band.
  2. Changing the polarization from vertical to horizontal.
  3. Lowering the transmit frequency while keeping the antenna efficiency acceptable. Correct answer.
  4. Increasing the antenna height above ground at both ends only.

Explanation

Ground-wave attenuation rises steeply with frequency because the induced currents in the lossy earth dissipate more energy as frequency increases. Lowering the frequency within the MF/HF band therefore extends the useful ground-wave range over land, provided the antenna remains efficient at the longer wavelength. Antenna height and polarization are secondary compared with the strong frequency dependence of ground-wave field strength.

Question 29Radio wave propagation and antennas

A satellite at 400 km altitude transmits at 145 MHz to a ground station. Which propagation mechanism dominates the path, and why?

  1. Sky-wave refraction from the F2 layer, because 145 MHz is below the critical frequency.
  2. Tropospheric ducting, because the satellite is inside the troposphere.
  3. Free-space line-of-sight propagation, because the ionosphere is essentially transparent at VHF and the path clears the horizon. Correct answer.
  4. Ground-wave propagation, because the signal follows the earth's surface.

Explanation

At 145 MHz the ionosphere is far above the critical frequency for typical conditions, so the wave passes through rather than being refracted back to earth. With the satellite well above the horizon, the path behaves as free-space line-of-sight, and only atmospheric absorption and scintillation add small losses. This is why satellite VHF links are computed with the free-space path loss formula.

Question 30Radio wave propagation and antennas

An HF circuit uses a single-hop F2 reflection with a fixed virtual reflection height. As the path length is increased from very short distances, why does the maximum usable frequency rise before eventually falling?

  1. Longer paths require a larger angle of incidence at the layer, so the same plasma density can refract higher frequencies back to earth up to the skip-distance limit. Correct answer.
  2. Longer paths increase the transmitted power density at the ionosphere, raising the plasma frequency.
  3. Longer paths lower the effective reflection height, which increases the critical frequency.
  4. Longer paths reduce the free-space loss so strongly that higher frequencies become receivable.

Explanation

The secant law relates the maximum usable frequency to the critical frequency multiplied by the secant of the incidence angle at the layer. A longer ground distance for a fixed layer height means a more oblique ray, a larger incidence angle and a larger secant factor, so higher frequencies can be returned. Beyond the geometry limit where the ray would no longer intersect the layer usefully, the advantage disappears and the MUF falls again.

Question 31RF circuits and components

A coaxial cable with a 50-ohm characteristic impedance is terminated by a purely resistive 50-ohm load. What is the voltage standing wave ratio on the line?

  1. Infinite
  2. 1:1 Correct answer.
  3. 50:1
  4. 2:1

Explanation

VSWR equals (1+|Γ|)/(1−|Γ|), where Γ is the load reflection coefficient. A load equal to the characteristic impedance gives Γ = 0, so the VSWR becomes exactly 1:1 and all incident power is absorbed by the load.

Question 32RF circuits and components

Which property of an ideal RF amplifier is described by its 1 dB compression point (P1dB)?

  1. The output power at which gain falls 1 dB below its small-signal linear value Correct answer.
  2. The frequency at which gain drops by 1 dB relative to midband
  3. The input power at which the noise figure increases by 1 dB
  4. The power level at which the amplifier becomes unconditionally stable

Explanation

P1dB is a large-signal linearity metric defined as the output (or input) power where the actual gain is 1 dB lower than the extrapolated small-signal gain. It marks the onset of gain compression as the device approaches saturation.

Question 33RF circuits and components

In a superheterodyne receiver, an RF signal at 100 MHz is mixed with a local oscillator at 90 MHz. What is the intermediate frequency produced by the difference product?

  1. 90 MHz
  2. 190 MHz
  3. 100 MHz
  4. 10 MHz Correct answer.

Explanation

The mixer produces sum and difference products of the RF and LO frequencies. The difference product is |100 − 90| = 10 MHz, which is the intermediate frequency. The sum product (190 MHz) is removed by filtering.

Question 34RF circuits and components

What is the primary function of an impedance matching network between an RF source and a load?

  1. To increase the overall noise figure of the chain
  2. To convert a balanced signal into an unbalanced one
  3. To double the operating frequency of the source
  4. To maximize power transfer and minimize reflections Correct answer.

Explanation

A matching network transforms the load impedance so that it equals the source impedance (or its complex conjugate), maximizing power transfer and minimizing reflected power. It may also shape bandwidth or filter harmonics, but its core purpose is impedance transformation.

Question 35RF circuits and components

Which component provides a DC return path while blocking RF energy from reaching a bias supply?

  1. A directional coupler
  2. An RF choke (inductor) Correct answer.
  3. A circulator
  4. A series coupling capacitor

Explanation

An RF choke presents high impedance at RF frequencies while offering a low-resistance DC path, isolating the bias supply from the RF signal. This is a standard bias-tee element in amplifier and mixer circuits.

Question 36RF circuits and components

What does a 3 dB power loss correspond to in terms of power ratio?

  1. A factor of about 3
  2. A factor of about 10
  3. A factor of about 2 Correct answer.
  4. A factor of about 1.5

Explanation

The decibel is defined as 10·log10(P2/P1) for power. A 3 dB change corresponds to a power ratio of approximately 2, since 10·log10(2) ≈ 3.01 dB.

Question 37RF circuits and components

Which characteristic impedance is most commonly used for coaxial cable in professional RF and test equipment?

  1. 600 ohms
  2. 50 ohms Correct answer.
  3. 300 ohms
  4. 75 ohms

Explanation

50 ohms is the standard for most RF systems because it balances low loss and high power-handling capability. It is used for test equipment, antennas, and general-purpose coaxial interconnects.

Question 38RF circuits and components

A filter passes frequencies from 1.8 GHz to 1.9 GHz and strongly attenuates all others. What type of filter is this?

  1. Low-pass
  2. High-pass
  3. Band-stop
  4. Band-pass Correct answer.

Explanation

A band-pass filter transmits energy within a defined range and rejects frequencies above and below it. The 1.8–1.9 GHz passband with attenuation outside matches this definition.

Question 39RF circuits and components

An amplifier has a gain of 20 dB and an input power of 1 mW. What is the output power?

  1. 10 mW
  2. 20 mW
  3. 200 mW
  4. 100 mW Correct answer.

Explanation

A 20 dB gain means the power ratio is 10^(20/10) = 100. Multiplying 1 mW by 100 gives 100 mW, which is also 20 dBm.

Question 40RF circuits and components

A receiver has a noise figure of 3 dB and a bandwidth of 1 MHz. What is the approximate thermal noise floor at its input at room temperature?

  1. −174 dBm
  2. −144 dBm
  3. −84 dBm
  4. −114 dBm Correct answer.

Explanation

The thermal noise floor is kTB, which at 290 K is about −174 dBm/Hz. Adding 10·log10(1 MHz) = 60 dB gives −114 dBm, and the 3 dB noise figure raises the referred noise to about −111 dBm, so −114 dBm is the closest listed value for the input floor.

Question 41RF circuits and components

In a 50-ohm system, a load of 100 ohms purely resistive is connected. What is the magnitude of the reflection coefficient?

  1. 0.67
  2. 1.00
  3. 0.50
  4. 0.33 Correct answer.

Explanation

The reflection coefficient magnitude is |Γ| = |(ZL − Z0)/(ZL + Z0)|. Substituting gives |(100 − 50)/(100 + 50)| = 50/150 ≈ 0.33.

Question 42RF circuits and components

What is the principal advantage of a double-balanced mixer over a single-ended mixer?

  1. It suppresses even-order products and provides port-to-port isolation Correct answer.
  2. It operates only at baseband frequencies
  3. It eliminates the need for any local oscillator
  4. It removes the need for an intermediate frequency stage

Explanation

A double-balanced mixer uses a balanced transformer and diode or transistor arrangement that cancels even-order mixing products and improves isolation between RF, LO, and IF ports. This reduces spurious responses in a receiver.

Question 43RF circuits and components

An RF filter shows a return loss of 20 dB at its center frequency. What does this indicate about the input match?

  1. All incident power is reflected
  2. The filter is completely unmatched
  3. About 20 percent of the incident power is reflected
  4. Only about 1 percent of the incident power is reflected Correct answer.

Explanation

Return loss in dB is −20·log10|Γ|, so 20 dB corresponds to |Γ| = 0.1. The reflected power fraction is |Γ|² = 0.01, meaning about 1 percent of incident power is reflected.

Question 44RF circuits and components

A quarter-wave transformer is used to match a 200-ohm load to a 50-ohm line. What characteristic impedance should the matching section have?

  1. 400 ohms
  2. 100 ohms Correct answer.
  3. 125 ohms
  4. 50 ohms

Explanation

For a quarter-wave transformer, Z_match = sqrt(Z0 · ZL). Taking sqrt(50 × 200) = sqrt(10000) = 100 ohms gives the required impedance.

Question 45RF circuits and components

What is the purpose of a circulator in an RF transmit/receive front end?

  1. To route the transmit signal to the antenna while directing the received signal to the receiver Correct answer.
  2. To amplify the received signal before the antenna
  3. To filter harmonics from the transmitter output
  4. To convert the RF signal to an intermediate frequency

Explanation

A circulator is a nonreciprocal three-port device that directs power from port to port in a defined rotation. In a T/R front end it lets the transmitter feed the antenna while isolating the receiver, and routes the received signal to the receiver.

Question 46RF circuits and components

Which phenomenon limits the dynamic range at the upper end of a receiver's usable input range?

  1. Phase noise of the reference oscillator
  2. Gain compression and intermodulation distortion Correct answer.
  3. Thermal noise at the antenna
  4. Skin effect in the antenna cable

Explanation

At high input levels, active stages compress and generate intermodulation products that corrupt the desired signal. This sets the upper limit of dynamic range, while thermal noise sets the lower limit.

Question 47RF circuits and components

A microstrip line is designed on a substrate with a relative permittivity of 4. What happens to the effective wavelength compared with free space?

  1. It becomes infinite
  2. It is unchanged
  3. It is shortened by roughly a factor of two Correct answer.
  4. It is lengthened by roughly a factor of two

Explanation

Wavelength on a dielectric-backed line is reduced by the square root of the effective permittivity. For εr = 4, the factor is about 2, so the guided wavelength is roughly half the free-space value.

Question 48RF circuits and components

What is the main reason for using a low-noise amplifier as the first stage in a receiver chain?

  1. It provides the highest output power in the chain
  2. It converts the RF signal directly to baseband
  3. The Friis formula shows that first-stage noise dominates the overall noise figure Correct answer.
  4. It eliminates the need for an image-reject filter

Explanation

According to the Friis formula, the noise contribution of each stage is divided by the gain of preceding stages. Placing a high-gain, low-noise stage first minimizes the overall system noise figure.

Question 49RF circuits and components

A directional coupler has a coupling factor of 20 dB and a directivity of 30 dB. What is the isolation between the input and isolated ports?

  1. 30 dB
  2. 20 dB
  3. 50 dB Correct answer.
  4. 10 dB

Explanation

Isolation equals coupling plus directivity for a directional coupler. Adding 20 dB coupling and 30 dB directivity gives 50 dB of isolation between the input and isolated ports.

Question 50RF circuits and components

Which statement correctly describes the S-parameter S21 of a two-port RF network?

  1. It is the output reflection coefficient at port 2
  2. It is the input reflection coefficient at port 1
  3. It is the forward transmission coefficient from port 1 to port 2 Correct answer.
  4. It is the reverse transmission coefficient from port 2 to port 1

Explanation

In S-parameter notation, S21 represents the wave transmitted from port 1 to port 2 with all other ports matched. Its magnitude squared gives the forward power gain of the network.