Exercises
ex10-01-ofdm-resolution
EasyA Wi-Fi access point at GHz transmits OFDM with subcarriers, kHz, and symbols. Compute the range resolution, maximum unambiguous range (assuming ), and velocity resolution.
.
Range resolution: .
Range resolution
MHz. m.
Maximum unambiguous range
m.
Velocity resolution
m. . m/s. The poor Doppler resolution is due to the very short CPI.
ex10-02-data-compensation
EasyExplain why dividing by is necessary in OFDM radar processing. What happens at subcarriers where (null subcarriers)?
The data symbols modulate the phase of each subcarrier.
Null subcarriers provide no radar illumination.
Data compensation
The received signal includes the data modulation as a multiplicative factor. Without compensation, the data acts as random phase rotations that destroy the 2D sinusoidal structure needed for coherent FFT processing. Division by restores the clean delay-Doppler phase structure.
Null subcarriers
Null subcarriers () provide no illumination and cannot be compensated. These entries are excluded from the measurement vector, creating a partial Fourier sensing matrix with missing rows. Compressed sensing methods can recover the missing information if the scene is sparse.
ex10-03-cp-range
EasyA 5G NR system with kHz (normal CP, ) is used for sensing. A second system uses kHz (). Compare the maximum unambiguous ranges and discuss the trade-off with range resolution.
Both systems can use the same total bandwidth , but with different numbers of subcarriers.
Max range for $\Delta f = 30$ kHz
m.
Max range for $\Delta f = 120$ kHz
m.
Trade-off
Smaller gives longer CP and larger , but the OFDM symbol duration is longer (), leading to worse Doppler resolution for a fixed number of symbols. Range resolution depends only on total bandwidth , which can be the same for both.
ex10-04-dirichlet-sidelobe
EasyThe Dirichlet kernel has peak sidelobes at dB. If you apply a Hamming window before the FFT, what is the expected peak sidelobe level? What is the cost in terms of main-lobe width?
Look up standard window properties.
Hamming window properties
Hamming window: peak sidelobe dB, main-lobe width that of the rectangular window.
Trade-off
Applying Hamming reduces sidelobes by about 30 dB but degrades range/Doppler resolution by a factor of . For a system with m (rectangular), the effective resolution becomes m with Hamming. This is the classic resolution-sidelobe trade-off.
ex10-05-otfs-vs-ofdm
EasyIn two sentences, state the key advantage of OTFS over OFDM for sensing fast-moving targets. What is the computational cost of converting between OFDM and OTFS representations?
Think about channel sparsity in the delay-Doppler domain.
The conversion involves a 2D DFT/IDFT.
Key advantage
In the OTFS delay-Doppler domain, each point target produces exactly one non-zero channel entry regardless of velocity, while in OFDM the high Doppler spreads each target's energy across many subcarriers via ICI. This sparsity makes OTFS robust to high-Doppler scenarios where OFDM fails.
Computational cost
Converting from delay-Doppler to time-frequency (or vice versa) requires an ISFFT/SFFT, which is a 2D-DFT of size . The cost is --- negligible compared to the baseband processing.
ex10-06-sensing-matrix-kronecker
MediumShow that the OFDM sensing matrix for the range-Doppler estimation problem can be written as a Kronecker product where and are (partial) DFT matrices. What are the dimensions of each factor?
The entry of depends on delay via only and on Doppler via only.
Use the property that .
Factor the steering vector
The -th column of is where () and ().
Assemble the Kronecker product
. If the targets are on a grid ( and ), then is the -th column of and is the -th column of . Dimensions: (or for an oversampled grid) and (or ).
ex10-07-ici-sir
MediumDerive the approximate SIR due to ICI in OFDM as a function of normalised Doppler . At what value of does the SIR drop below 10 dB?
Start from the DFT of over .
Sum the ICI power from all interfering subcarriers.
ICI power calculation
The DFT output on subcarrier for a target at normalised Doppler is . The desired signal power is (for ). The ICI power sums to for small .
SIR expression
.
Find the 10 dB threshold
dB gives , so . At , the ICI-induced SIR drops below 10 dB.
ex10-08-zak-properties
MediumProve the quasi-periodicity property of the Zak transform: .
Substitute into the Zak transform definition and re-index the sum.
Substitute and re-index
. Let , so and :
Recognise the Zak transform
The remaining sum is exactly , proving .
ex10-09-otfs-circular-convolution
MediumExplain why the OTFS input-output relation is a 2D circular convolution rather than a linear convolution. What role does the cyclic prefix play?
The CP converts linear convolution to circular convolution in the delay dimension.
The finite DFT aperture in slow-time converts to circular in the Doppler dimension.
Delay dimension
The cyclic prefix of each OFDM symbol ensures that the delay-domain convolution between the channel and the transmitted signal is circular (modulo delay bins). Without CP, the convolution would be linear and the OFDM symbols would suffer ISI.
Doppler dimension
The SFFT operates over a finite window of OFDM symbols. The Doppler processing via DFT implicitly assumes periodicity in the slow-time dimension, making the Doppler convolution circular (modulo Doppler bins). This is the standard DFT circular convolution property.
Combined
Together, the CP (delay) and finite DFT window (Doppler) make the full 2D channel convolution circular. This circularity is essential for the block-circulant structure of the OTFS sensing matrix and the efficiency of FFT-based algorithms.
ex10-10-fractional-doppler
MediumFor an OTFS system with Doppler bins and a target with fractional Doppler , compute the fraction of energy that leaks to adjacent Doppler bins. How many bins contain at least 1% of the total energy?
The leakage follows a Dirichlet kernel with parameter and offset .
Energy in the main bin
The energy in the nearest bin () is proportional to . For large : . About 52% of energy is in the main bin.
Leakage to adjacent bins
The -th adjacent bin has energy for large . For : and . Bins with energy: computing gives roughly , so about 7 bins.
ex10-11-ambiguity-volume
MediumThe ambiguity function satisfies the volume constraint (for unit-energy waveforms). What does this imply about the trade-off between main-lobe width and sidelobe level?
The total volume under is fixed.
A narrower main lobe must be compensated by higher sidelobes elsewhere.
Volume conservation
Since the total volume is fixed at 1, any reduction in main-lobe width (better resolution) must be compensated by redistributing volume to the sidelobes. A thumbtack ambiguity function (like OFDM with random data) spreads the sidelobes thinly across the entire delay-Doppler plane.
Implications
No waveform can have both arbitrarily narrow main lobe and arbitrarily low sidelobes. OFDM with random data achieves a good compromise: the sidelobes are at dB but spread uniformly (like white noise). Windowing reduces peak sidelobes but widens the main lobe, keeping the volume constant. This is the Woodward ambiguity principle.
ex10-12-fmcw-vs-ofdm
HardAn automotive radar at 77 GHz has GHz. Compare the range-Doppler maps obtained by FMCW (1024-sample chirp, 128 chirps) and OFDM (, , kHz) for a scene with two targets at and . Which waveform resolves both targets?
Compute range resolution for both: .
Check if the Doppler shift causes ICI in OFDM.
Range resolution
Both: m. The targets are separated by 0.1 m , so neither waveform resolves them in range. However, they differ in velocity.
Doppler shift
kHz. Normalized: .
FMCW performance
FMCW has no ICI issue. The two targets overlap in range but are separated in Doppler. With chirps and PRI , m/s. Target 2 is at 22.2 m/s. Both targets are visible, separated by Doppler bins.
OFDM performance
With , ICI is negligible ( dB). OFDM also resolves both targets via the Doppler dimension. Performance is comparable to FMCW in this low-Doppler regime.
ex10-13-otfs-sensing-matrix
HardDerive the block-circulant structure of the OTFS sensing matrix. Starting from the 2D circular convolution , show that the vectorised input-output relation has as a block-circulant matrix with circulant blocks.
Vectorise using lexicographic ordering: .
A 1D circular convolution gives a circulant matrix. The 2D case nests two levels.
Vectorise
Define and similarly for and . The index mapping is for , .
Block structure
Partition into blocks of size . Block of is where is an circulant matrix with first column given by . This is a block-circulant matrix (the block index depends only on ).
Circulant blocks
Each block is itself circulant because the delay convolution is circular (modulo ). Hence is block-circulant with circulant blocks (BCCB), which is diagonalised by --- the 2D-DFT.
ex10-14-papr-comparison
HardThe PAPR of an OFDM signal with subcarriers and i.i.d. QPSK symbols is approximately dB with high probability. For , compute the PAPR. Compare with FMCW (constant envelope) and discuss the impact on PA efficiency and sensing range.
PA efficiency degrades with back-off. At 10 dB back-off, a class-A PA has efficiency.
OFDM PAPR
dB. In practice, with clipping at CCDF, PAPR -- dB.
PA back-off
With 10 dB PAPR, the PA must operate at 10 dB back-off. If the PA has 1 W saturated output, the average output is 0.1 W. For FMCW (PAPR dB), the full 1 W is available.
Impact on sensing range
The radar range equation gives . A 10 dB reduction in average power reduces maximum detection range by factor . FMCW can detect targets farther than OFDM with the same PA. This is a significant disadvantage for OFDM/OTFS in automotive radar.
ex10-15-sensing-matrix-coherence
HardThe coherence of a sensing matrix is . For the OFDM sensing matrix with targets on a grid, show that depends on the minimum delay-Doppler separation. What is the coherence for two targets separated by exactly one resolution cell?
The columns of are Kronecker products of DFT vectors.
The inner product of two DFT columns is a Dirichlet kernel.
Inner product of two columns
. Each factor is a Dirichlet kernel evaluated at the delay or Doppler separation.
One-resolution-cell separation
If (one range bin) and , then (orthogonal) and . So --- targets separated by exactly one grid cell are orthogonal.
Sub-grid separation
For targets separated by a fraction of the resolution cell, , which approaches 1 as . High coherence makes sparse recovery difficult for closely spaced targets.
ex10-16-pmcw-sidelobes
HardA PMCW radar uses an m-sequence of length . Compute the peak sidelobe level in the range dimension. Compare with OFDM (, no window) and discuss implications for weak target detection.
The autocorrelation of an m-sequence of length has peak sidelobe .
M-sequence sidelobe
The periodic autocorrelation of an m-sequence of length is for and for . The peak sidelobe level is dB.
OFDM sidelobe
OFDM with rectangular window: Dirichlet kernel with peak sidelobe dB.
Comparison
PMCW with m-sequence has dB better sidelobe suppression than unwindowed OFDM. For detecting a weak target ( dB below a strong target), the OFDM sidelobes at dB would mask it, but PMCW sidelobes at dB would not. The OFDM system would need Hamming windowing ( dB) to match PMCW performance.
ex10-17-isac-rate-sensing
ChallengeAn OFDM ISAC system allocates subcarriers to pilot symbols (for sensing) and to data symbols (for communication). Derive the sensing SNR and communication rate as functions of , and find the Pareto-optimal trade-off curve. Assume total transmit power , flat-fading channel with gain , and noise variance .
Sensing SNR scales with the number of pilot subcarriers.
Communication rate scales with the number of data subcarriers.
Sensing SNR
With pilot subcarriers, the sensing observation has measurements. The per-measurement SNR is (power per subcarrier). The effective sensing SNR after matched filtering is .
Communication rate
With data subcarriers, the ergodic rate is bits/s.
Pareto curve
The pair is parametrised by . As increases, sensing SNR increases linearly while communication rate decreases linearly. The Pareto frontier is the line segment connecting to . The optimal operating point depends on the relative importance of sensing and communication.
ex10-18-otfs-crb
ChallengeDerive the Cramer-Rao bound for estimating the delay and Doppler of a single target using OTFS with subcarriers and symbols. Show that the CRBs match those of a dedicated radar waveform with the same bandwidth and CPI.
The Fisher information matrix for involves derivatives of the observation w.r.t. these parameters.
For a single complex sinusoid in AWGN, the CRB depends on bandwidth and observation time.
Signal model
After OTFS demodulation, the observation for a single target is where is concentrated near . In the time-frequency domain (equivalent), .
Fisher information
The FIM for is where is the noise-free signal. .
CRBs
For large and : . Similarly, . These are the standard CRBs for frequency estimation from samples of a complex sinusoid --- identical to what a dedicated radar waveform achieves with the same bandwidth and CPI.
ex10-19-multi-target-sensing-matrix
ChallengeConsider an OFDM sensing system with , , and targets at arbitrary (off-grid) delay-Doppler locations. The scene is discretised onto a -oversampled grid (, ). Write the compressed sensing formulation, identify the sensing matrix , and discuss the RIP properties of this matrix.
The sensing matrix is a partial Fourier matrix on an oversampled grid.
Partial Fourier matrices satisfy RIP with high probability when the number of measurements exceeds .
CS formulation
where (vectorised to ), (vectorised to ) is -sparse with , and is a partial 2D DFT evaluated on the oversampled grid.
RIP analysis
The sensing matrix where selects the measurement rows. Partial Fourier matrices satisfy -RIP with constant when . Since , we need , which our measurements do NOT satisfy.
Practical implication
The RIP bound is pessimistic for structured (Fourier) matrices. In practice, OFDM sensing with measurements per target works well for sparse recovery (LASSO, OAMP) even though the theoretical RIP bound is not met. The oversampled grid allows sub-resolution target localisation.
ex10-20-waveform-a-structure
ChallengeFor each of the four waveforms (FMCW, OFDM, OTFS, PMCW), write the sensing matrix in terms of structured matrix factors (DFT, circulant, diagonal, permutation). For each, state the cost of a matrix-vector product and in terms of , , and (number of grid points). These costs determine the computational feasibility of iterative recovery algorithms.
Exploit Kronecker and FFT structure.
Block-circulant matrices are diagonalised by 2D-DFT.
FMCW
(two DFT factors). Cost: where .
OFDM
(partial Kronecker DFT). Cost: via row-column FFT. is a row selection (free).
OTFS
where is BCCB. Cost: via 2D-FFT to diagonalise the BCCB part, plus element-wise multiplication with .
PMCW
where is circulant (code shifts). Cost: since circulant products use FFT. All four waveforms support matrix-vector products, making iterative sparse recovery feasible for scenes with .