Electromagnetic Waves, Antennas, and RF: RF Interference and Jamming - Exercises
ElectromagneticWavesAntennasAndRFRFInterferenceAndJammingExercises
2026-10-10 04:01:28
2026-10-10 04:01:28
Example
Electromagnetic Waves, Antennas, and RF: RF Interference and Jamming
added synonym for series
RF interference
radio-frequency interference
interference-to-signal ratio
J/S
J/N0
J0/N0
admitted interference power
spectral overlap
equivalent noise bandwidth
C/(N+J)
receiver degradation
interference temperature
antenna discrimination
interference margin
GNSS interference
exercises
worked solutions
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\section*{Electromagnetic Waves, Antennas, and RF:\ RF Interference and Jamming - Exercises and Complete Worked Solutions}
This companion to EM27 develops calculation skill with receiver-side interference metrics. The problems move from basic power ratios to spectral overlap, equivalent noise bandwidth, receiver degradation, antenna discrimination, equivalent interference temperature, and a final GNSS-style allowable-interference calculation. The emphasis is defensive receiver analysis: all powers are referred to a common receiver plane and the limits of the additive-power model are stated explicitly \cite{Sklar2001,Pozar2012,KaplanHegarty2017,Betz2016}.
The principal relations are
\begin{equation}
\left(\frac{J}{S}\right)_{\rm dB}=J_{\rm dBW}-S_{\rm dBW},
\end{equation}
\begin{equation}
\left(\frac{J}{N_0}\right)_{\rm dB-Hz}=J_{\rm dBW}-N_{0,\rm dBW/Hz},
\end{equation}
\begin{equation}
\frac{J}{N}=\frac{J/N_0}{B_n},
\end{equation}
\begin{equation}
\frac{C}{N+J}=\frac{C/N}{1+J/N},
\end{equation}
and
\begin{equation}
D_{\rm dB}=10\log_{10}\left(1+\frac{J}{N}\right).
\end{equation}
For flat noise-like interference,
\begin{equation}
\frac{C}{N_0+J_0}=\frac{C/N_0}{1+J_0/N_0}.
\end{equation}
Use
\begin{equation}
k=1.380649\times10^{-23}\,\text{J/K}.
\end{equation}
\begin{center}
\includegraphics[width=0.84\textwidth,keepaspectratio]{EM27E1_fig01_metric_conversion_map.png}
{\small Figure 1. The main conversions used throughout the exercises. Total powers, spectral densities, and finite-bandwidth powers must not be mixed without the appropriate bandwidth conversion.}
\end{center}
\section*{Part I: Exercises}
\subsection*{Exercise 1: interference-to-signal ratio}
At a common receiver reference plane,
\begin{equation}
S=-128\,\text{dBm},\qquad J=-136\,\text{dBm}.
\end{equation}
Find $J/S$ in dB and linear form.
\subsection*{Exercise 2: $J/N_0$ from system temperature}
A receiver has
\begin{equation}
T_{\rm sys}=320\,\text{K}
\end{equation}
and admits an interference power
\begin{equation}
J=-148\,\text{dBW}.
\end{equation}
Find $N_0$ in dBW/Hz and $J/N_0$ in dB-Hz.
\subsection*{Exercise 3: convert $C/N_0$ and $J/N_0$ to $J/S$}
At the same reference plane,
\begin{equation}
C/N_0=47\,\text{dB-Hz},\qquad J/N_0=39\,\text{dB-Hz}.
\end{equation}
Assuming $S=C$, find $J/S$.
\subsection*{Exercise 4: bandwidth conversion from $J/N_0$ to $J/N$}
A receiver has
\begin{equation}
J/N_0=53\,\text{dB-Hz},\qquad B_n=200\,\text{kHz}.
\end{equation}
Find $J/N$ in dB and linear form.
\subsection*{Exercise 5: additive interference degradation}
A noise-only receiver has
\begin{equation}
C/N=12\,\text{dB}.
\end{equation}
An additive in-band interferer produces
\begin{equation}
J/N=-3\,\text{dB}.
\end{equation}
Find the degradation $D_{\rm dB}$ and the resulting $C/(N+J)$.
\subsection*{Exercise 6: invert a degradation requirement}
Define an \emph{interference degradation budget} $D_{\max}$ as the largest permitted additive degradation. For
\begin{equation}
D_{\max}=1.00\,\text{dB},
\end{equation}
find the largest allowable $J/N$ in linear form and in dB.
\subsection*{Exercise 7: broadband spectral overlap}
A noise-like interferer has total received power
\begin{equation}
J_{\rm total}=-92\,\text{dBm}
\end{equation}
uniformly distributed over $16\,\text{MHz}$. An ideal rectangular receiver admits $2\,\text{MHz}$ entirely inside that band. Assume $T=290\,\text{K}$. Find:
\begin{enumerate}
\item admitted interference power $J_{\rm in}$;
\item thermal-noise power $N$;
\item $J/N$;
\item additive degradation $D_{\rm dB}$.
\end{enumerate}
\begin{center}
\includegraphics[width=0.78\textwidth,keepaspectratio]{EM27E1_fig02_spectral_overlap_fraction.png}
{\small Figure 2. For a uniform broadband interferer and an ideal rectangular receiver, the admitted fraction equals the overlap bandwidth divided by the interference bandwidth.}
\end{center}
\subsection*{Exercise 8: equivalent noise bandwidth of a first-order low-pass filter}
A one-sided normalized receiver power response is
\begin{equation}
|H(f)|^2=\frac{1}{1+(f/f_c)^2},\qquad f\ge0,
\end{equation}
with
\begin{equation}
f_c=100\,\text{kHz}.
\end{equation}
Find the one-sided equivalent noise bandwidth
\begin{equation}
B_n=\int_0^\infty |H(f)|^2\,df.
\end{equation}
If a flat interference density is
\begin{equation}
J_0=-175\,\text{dBm/Hz},
\end{equation}
find the admitted interference power and, at $290\,\text{K}$, the density-domain degradation.
\subsection*{Exercise 9: noise-like interference density}
A receiver has
\begin{equation}
C/N_0=46.5\,\text{dB-Hz}
\end{equation}
and a flat noise-like interference density satisfying
\begin{equation}
J_0/N_0=-8\,\text{dB}.
\end{equation}
Find the degradation and the effective $C/(N_0+J_0)$.
\subsection*{Exercise 10: equivalent interference temperature}
A receiver has
\begin{equation}
T_{\rm sys}=350\,\text{K}.
\end{equation}
It admits
\begin{equation}
J=2.0\times10^{-15}\,\text{W}
\end{equation}
over
\begin{equation}
B_n=2.0\,\text{MHz}.
\end{equation}
Find the equivalent interference temperature $T_J$, the effective temperature $T_{\rm eff}$, and the corresponding degradation.
\subsection*{Exercise 11: two independent RF links}
A desired transmitter and an interfering transmitter terminate at the same receiver. Their link terms are
\begin{align}
\text{EIRP}_S&=20\,\text{dBW}, & L_{S,\rm path}&=140\,\text{dB}, & L_{S,\rm other}&=2\,\text{dB}, & G_r(\Omega_S)&=6\,\text{dBi},\\
\text{EIRP}_J&=10\,\text{dBW}, & L_{J,\rm path}&=132\,\text{dB}, & L_{J,\rm other}&=1\,\text{dB}, & G_r(\Omega_J)&=-4\,\text{dBi}.
\end{align}
Find $S$, $J$, and $J/S$ in dB.
\subsection*{Exercise 12: receive-antenna interference discrimination}
Before spatial discrimination, an interfering signal has
\begin{equation}
J/S=-3\,\text{dB}.
\end{equation}
A directional receive antenna reduces the gain toward the interferer by $18\,\text{dB}$ relative to the desired direction, with all other terms unchanged. Find the new $J/S$. State how $J/N_0$ changes if the receiver noise density at the chosen reference plane is unchanged.
\begin{center}
\includegraphics[width=0.76\textwidth,keepaspectratio]{EM27E1_fig04_two_link_antenna_discrimination.png}
{\small Figure 3. Desired and interfering signals are separate links. Receive-antenna discrimination can change the interference power without changing the desired-signal path.}
\end{center}
\subsection*{Exercise 13: combine $C/N_0$, $J/N_0$, and finite bandwidth}
Let
\begin{equation}
C/N_0=60\,\text{dB-Hz},\qquad J/N_0=55\,\text{dB-Hz},\qquad B_n=100\,\text{kHz}.
\end{equation}
Find $C/N$, $J/N$, the degradation, and $C/(N+J)$.
\subsection*{Exercise 14: same total interference power, different overlap}
Two broadband interferers each arrive with total power
\begin{equation}
J_{\rm total}=-90\,\text{dBm}
\end{equation}
uniformly spread over $20\,\text{MHz}$. For interferer A, a $2\,\text{MHz}$ receiver lies fully inside the interference band. For interferer B, only $0.5\,\text{MHz}$ of the receiver passband overlaps the interference band. Find the admitted interference powers and their difference in dB.
\subsection*{Exercise 15: GNSS-style density-domain degradation}
An illustrative receiver has
\begin{equation}
C/N_0=47.08\,\text{dB-Hz},\qquad J_0/N_0=-6.0\,\text{dB}.
\end{equation}
For
\begin{equation}
B_n=2.0\,\text{MHz},
\end{equation}
find the density-domain degradation, the effective $C/(N_0+J_0)$, and the corresponding finite-bandwidth $C/(N+J)$.
\subsection*{Exercise 16: allowable noise-like interference from a receiver requirement}
The same illustrative receiver has nominal
\begin{equation}
C/N_0=47.08\,\text{dB-Hz}
\end{equation}
and must maintain
\begin{equation}
C/(N_0+J_0)\ge42.00\,\text{dB-Hz}.
\end{equation}
Assuming additive uncorrelated noise-like interference, determine the maximum allowable $J_0/N_0$ in linear form and in dB.
\begin{center}
\includegraphics[width=0.80\textwidth,keepaspectratio]{EM27E1_fig03_degradation_budget_curve.png}
{\small Figure 4. Additive degradation as a function of $J/N$ or, for flat noise-like interference, $J_0/N_0$. The inverse relation converts a degradation budget into an allowable interference ratio.}
\end{center}
\section*{Part II: Complete Worked Solutions}
\subsection*{Solution 1: interference-to-signal ratio}
Because both powers are expressed in the same logarithmic units at the same reference plane,
\begin{align}
(J/S)_{\rm dB}
&=J_{\rm dBm}-S_{\rm dBm}\\
&=-136-(-128)\\
&=-8.00\,\text{dB}.
\end{align}
The linear ratio is
\begin{equation}
\frac{J}{S}=10^{-8/10}=0.1585.
\end{equation}
Thus the interference power is about $15.8\%$ of the desired-signal power.
\subsection*{Solution 2: $J/N_0$ from system temperature}
The thermal-noise density is
\begin{equation}
N_0=kT_{\rm sys}.
\end{equation}
In dBW/Hz,
\begin{align}
N_{0,\rm dBW/Hz}
&=10\log_{10}\left[(1.380649\times10^{-23})(320)\right]\\
&=-203.55\,\text{dBW/Hz}.
\end{align}
Therefore
\begin{align}
(J/N_0)_{\rm dB-Hz}
&=-148-(-203.55)\\
&=55.55\,\text{dB-Hz}.
\end{align}
\subsection*{Solution 3: convert $C/N_0$ and $J/N_0$ to $J/S$}
With $S=C$,
\begin{equation}
\left(\frac{J}{S}\right)_{\rm dB}
=\left(\frac{J}{N_0}\right)_{\rm dB-Hz}
-\left(\frac{C}{N_0}\right)_{\rm dB-Hz}.
\end{equation}
Hence
\begin{equation}
J/S=39-47=-8.00\,\text{dB}.
\end{equation}
The dB-Hz units cancel because both ratios use the same $N_0$.
\subsection*{Solution 4: bandwidth conversion from $J/N_0$ to $J/N$}
The bandwidth term is
\begin{align}
10\log_{10}B_n
&=10\log_{10}(2.00\times10^5)\\
&=53.0103\,\text{dB-Hz}.
\end{align}
Therefore
\begin{align}
(J/N)_{\rm dB}
&=53.00-53.0103\\
&=-0.0103\,\text{dB}.
\end{align}
In linear form,
\begin{equation}
J/N=10^{-0.0103/10}=0.9976.
\end{equation}
The interference and thermal-noise powers are therefore almost equal.
\subsection*{Solution 5: additive interference degradation}
First convert the interference ratio:
\begin{equation}
J/N=10^{-3/10}=0.5012.
\end{equation}
The degradation is
\begin{align}
D_{\rm dB}
&=10\log_{10}(1+0.5012)\\
&=1.764\,\text{dB}.
\end{align}
Therefore
\begin{align}
C/(N+J)_{\rm dB}
&=(C/N)_{\rm dB}-D_{\rm dB}\\
&=12.00-1.764\\
&=10.236\,\text{dB}.
\end{align}
\subsection*{Solution 6: invert a degradation requirement}
Starting from
\begin{equation}
D_{\max}=10\log_{10}(1+J/N),
\end{equation}
exponentiate both sides:
\begin{equation}
10^{D_{\max}/10}=1+J/N.
\end{equation}
Hence the largest allowable ratio is
\begin{equation}
\left(\frac{J}{N}\right)_{\max}=10^{D_{\max}/10}-1.
\end{equation}
For $D_{\max}=1.00\,\text{dB}$,
\begin{equation}
(J/N)_{\max}=10^{0.1}-1=0.2589.
\end{equation}
In decibels,
\begin{equation}
(J/N)_{\max,\rm dB}=10\log_{10}(0.2589)=-5.87\,\text{dB}.
\end{equation}
\subsection*{Solution 7: broadband spectral overlap}
For a uniform spectrum, define the \emph{spectral-overlap fraction}
\begin{equation}
\eta_{\rm ov}=\frac{J_{\rm in}}{J_{\rm total}}.
\end{equation}
With complete containment of the receiver band,
\begin{equation}
\eta_{\rm ov}=\frac{2}{16}=0.125.
\end{equation}
Thus
\begin{align}
J_{\rm in,dBm}
&=-92+10\log_{10}(0.125)\\
&=-101.03\,\text{dBm}.
\end{align}
At $290\,\text{K}$,
\begin{equation}
N_{0,\rm dBm/Hz}\approx-173.98\,\text{dBm/Hz}.
\end{equation}
Therefore
\begin{align}
N_{\rm dBm}
&=-173.98+10\log_{10}(2.0\times10^6)\\
&=-110.96\,\text{dBm}.
\end{align}
The ratio is
\begin{equation}
(J/N)_{\rm dB}=-101.03-(-110.96)=9.93\,\text{dB}.
\end{equation}
The degradation is
\begin{align}
D_{\rm dB}
&=10\log_{10}\left(1+10^{9.93/10}\right)\\
&=10.35\,\text{dB}.
\end{align}
\subsection*{Solution 8: equivalent noise bandwidth of a first-order low-pass filter}
The one-sided equivalent noise bandwidth is
\begin{align}
B_n
&=\int_0^\infty \frac{df}{1+(f/f_c)^2}.
\end{align}
Let $u=f/f_c$, so $df=f_c\,du$. Then
\begin{align}
B_n
&=f_c\int_0^\infty\frac{du}{1+u^2}\\
&=f_c\left[\tan^{-1}u\right]_0^\infty\\
&=\frac{\pi}{2}f_c.
\end{align}
For $f_c=100\,\text{kHz}$,
\begin{equation}
B_n=157.08\,\text{kHz}.
\end{equation}
The admitted flat-spectrum interference is
\begin{align}
J_{\rm in,dBm}
&=J_{0,\rm dBm/Hz}+10\log_{10}B_n\\
&=-175+10\log_{10}(1.5708\times10^5)\\
&=-123.04\,\text{dBm}.
\end{align}
At $290\,\text{K}$,
\begin{equation}
N_0\approx-173.98\,\text{dBm/Hz},
\end{equation}
so
\begin{equation}
(J_0/N_0)_{\rm dB}=-175-(-173.98)=-1.02\,\text{dB}.
\end{equation}
The density-domain degradation is therefore
\begin{equation}
D_{\rm dB}=10\log_{10}\left(1+10^{-1.02/10}\right)=2.53\,\text{dB}.
\end{equation}
\subsection*{Solution 9: noise-like interference density}
Convert the density ratio to linear form:
\begin{equation}
J_0/N_0=10^{-8/10}=0.1585.
\end{equation}
Then
\begin{align}
D_{\rm dB}
&=10\log_{10}(1+0.1585)\\
&=0.639\,\text{dB}.
\end{align}
The effective density ratio is
\begin{align}
C/(N_0+J_0)_{\rm dB-Hz}
&=46.5-0.639\\
&=45.86\,\text{dB-Hz}.
\end{align}
\subsection*{Solution 10: equivalent interference temperature}
By definition,
\begin{equation}
T_J=\frac{J}{kB_n}.
\end{equation}
Thus
\begin{align}
T_J
&=\frac{2.0\times10^{-15}}
{(1.380649\times10^{-23})(2.0\times10^6)}\\
&=72.43\,\text{K}.
\end{align}
The effective temperature is
\begin{equation}
T_{\rm eff}=350+72.43=422.43\,\text{K}.
\end{equation}
The degradation is
\begin{align}
D_{\rm dB}
&=10\log_{10}\left(\frac{422.43}{350}\right)\\
&=0.817\,\text{dB}.
\end{align}
\subsection*{Solution 11: two independent RF links}
For the desired link,
\begin{align}
S_{\rm dBW}
&=20-140-2+6\\
&=-116\,\text{dBW}.
\end{align}
For the interfering link,
\begin{align}
J_{\rm dBW}
&=10-132-1-4\\
&=-127\,\text{dBW}.
\end{align}
Therefore
\begin{equation}
(J/S)_{\rm dB}=-127-(-116)=-11\,\text{dB}.
\end{equation}
The result shows why interference analysis is naturally a two-link problem: each source has its own EIRP, path loss, losses, and receive-antenna gain.
\subsection*{Solution 12: receive-antenna interference discrimination}
An $18\,\text{dB}$ reduction in receive gain toward the interferer reduces $J$ by $18\,\text{dB}$ while leaving $S$ unchanged. Therefore
\begin{equation}
(J/S)_{\rm new}=-3-18=-21\,\text{dB}.
\end{equation}
If $N_0$ at the chosen reference plane is unchanged, then $J/N_0$ also decreases by $18\,\text{dB}$. This statement assumes the antenna discrimination changes the admitted interference power without changing the receiver noise reference used for the comparison.
\subsection*{Solution 13: combine $C/N_0$, $J/N_0$, and finite bandwidth}
Since
\begin{equation}
10\log_{10}(100\,\text{kHz})=50\,\text{dB-Hz},
\end{equation}
we obtain
\begin{equation}
C/N=60-50=10\,\text{dB}
\end{equation}
and
\begin{equation}
J/N=55-50=5\,\text{dB}.
\end{equation}
Thus
\begin{equation}
J/N=10^{5/10}=3.162.
\end{equation}
The degradation is
\begin{equation}
D_{\rm dB}=10\log_{10}(1+3.162)=6.19\,\text{dB}.
\end{equation}
Therefore
\begin{equation}
C/(N+J)=10-6.19=3.81\,\text{dB}.
\end{equation}
\subsection*{Solution 14: same total interference power, different overlap}
For interferer A,
\begin{equation}
\eta_{\rm ov,A}=\frac{2}{20}=0.10,
\end{equation}
so
\begin{equation}
J_{\rm in,A}=-90+10\log_{10}(0.10)=-100.00\,\text{dBm}.
\end{equation}
For interferer B,
\begin{equation}
\eta_{\rm ov,B}=\frac{0.5}{20}=0.025,
\end{equation}
so
\begin{equation}
J_{\rm in,B}=-90+10\log_{10}(0.025)=-106.02\,\text{dBm}.
\end{equation}
The second receiver admits
\begin{equation}
-106.02-(-100.00)=-6.02\,\text{dB}
\end{equation}
less interference power. Equal total incident power therefore does not imply equal receiver degradation.
\subsection*{Solution 15: GNSS-style density-domain degradation}
The linear density ratio is
\begin{equation}
J_0/N_0=10^{-6/10}=0.2512.
\end{equation}
Therefore
\begin{equation}
D_{\rm dB}=10\log_{10}(1+0.2512)=0.973\,\text{dB}.
\end{equation}
The effective carrier-to-noise-plus-interference density ratio is
\begin{align}
C/(N_0+J_0)_{\rm dB-Hz}
&=47.08-0.973\\
&=46.11\,\text{dB-Hz}.
\end{align}
For $B_n=2.0\,\text{MHz}$,
\begin{equation}
10\log_{10}B_n=63.010\,\text{dB-Hz},
\end{equation}
so
\begin{align}
C/(N+J)_{\rm dB}
&=46.11-63.010\\
&=-16.90\,\text{dB}.
\end{align}
The negative finite-bandwidth ratio is not contradictory; spread-spectrum and correlation receivers can operate with negative pre-correlation $C/N$.
\subsection*{Solution 16: allowable noise-like interference from a receiver requirement}
The available degradation budget is
\begin{align}
D_{\max}
&=47.08-42.00\\
&=5.08\,\text{dB}.
\end{align}
For additive noise-like interference,
\begin{equation}
D_{\max}=10\log_{10}(1+J_0/N_0).
\end{equation}
Solving for the maximum allowable density ratio gives
\begin{equation}
\left(\frac{J_0}{N_0}\right)_{\max}
=10^{D_{\max}/10}-1.
\end{equation}
Therefore
\begin{equation}
\left(\frac{J_0}{N_0}\right)_{\max}
=10^{5.08/10}-1
=2.221.
\end{equation}
In decibels,
\begin{equation}
\left(\frac{J_0}{N_0}\right)_{\max,\rm dB}
=10\log_{10}(2.221)
=3.47\,\text{dB}.
\end{equation}
This is an allowable-interference result only for the additive, uncorrelated, noise-like model. Narrowband, pulsed, swept, correlated, or nonlinear front-end effects require receiver-specific analysis.
\section*{Summary of problem-solving strategy}
For receiver-side interference calculations, use the following sequence:
\begin{enumerate}
\item refer desired signal, interference, and noise quantities to a common receiver plane;
\item keep total powers distinct from spectral densities;
\item calculate admitted interference using spectral overlap or the actual transfer function;
\item convert density ratios to finite-bandwidth ratios only after specifying $B_n$;
\item apply the additive degradation formula only when that model is physically justified; and
\item compare the resulting metric with the receiver requirement or degradation budget.
\end{enumerate}
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\emph{Microwave Engineering},
4th ed., Wiley, 2012.
\bibitem{Balanis2016}
C. A. Balanis,
\emph{Antenna Theory: Analysis and Design},
4th ed., Wiley, 2016.
\bibitem{KaplanHegarty2017}
E. D. Kaplan and C. J. Hegarty, eds.,
\emph{Understanding GPS/GNSS: Principles and Applications},
3rd ed., Artech House, 2017.
\bibitem{Betz2016}
J. W. Betz,
\emph{Engineering Satellite-Based Navigation and Timing: Global Navigation Satellite Systems, Signals, and Receivers},
Wiley-IEEE Press, 2016.
\end{thebibliography}