Exercises
ex01-virtual-count
EasyA MIMO radar has transmit and receive antennas.
(a) How many virtual array elements does it create?
(b) What is the aperture efficiency ?
(c) If instead (same total ), what are and ? Which split is better for imaging?
.
Aperture efficiency is maximised when .
Part (a)
virtual elements.
Part (b)
.
Part (c)
virtual elements. . The equal split produces more virtual elements (64 vs. 60) and higher efficiency, confirming the AM-GM optimality.
ex02-ula-spacing
EasyDesign a MIMO ULA at GHz with and that creates a filled virtual array at spacing.
(a) What are the Tx and Rx element spacings?
(b) What is the total virtual array length?
(c) What is the angular resolution (first null)?
For a filled virtual ULA, one array has spacing and the other .
Element spacings
mm, so mm. Rx spacing: mm. Tx spacing: mm.
Virtual array length
elements. Length: mm.
Angular resolution
rad .
ex03-waveform-types
EasyList three methods for achieving waveform orthogonality in MIMO radar (TDM, FDM, CDM). For each, state:
(a) What resource is divided among transmitters.
(b) One advantage and one disadvantage relative to the others.
(c) Which is most commonly used in automotive radar and why.
TDM divides time; FDM divides frequency; CDM divides code space.
TDM
Divides time. Advantage: simplest implementation. Disadvantage: reduces effective CPI by , degrading Doppler resolution.
FDM
Divides frequency. Advantage: no cross-talk between Tx. Disadvantage: reduces per-Tx bandwidth, degrading range resolution.
CDM
Divides code space. Advantage: each Tx uses full bandwidth and full CPI. Disadvantage: requires good code design; residual cross-correlation creates sidelobes.
Automotive
TDM is most common in automotive radar due to its simplicity and low hardware cost. The reduced CPI is acceptable because automotive scenarios have short coherence times.
ex04-diversity-gain
MediumA distributed MIMO radar has independent Tx-Rx pairs detecting a Swerling-I target. The single-pair false-alarm probability is and the single-pair detection probability at the design SNR is .
(a) Compute the overall detection probability with square-law combining.
(b) How many pairs are needed to achieve ?
(c) What happens if the scattering cross-sections are correlated (Swerling-II instead of Swerling-I)?
for independent observations.
Correlated observations reduce the effective diversity order.
Part (a)
.
Part (b)
Need , so . Hence pairs.
Part (c)
Correlated scattering cross-sections reduce the effective diversity order below . In the extreme (fully correlated), the effective diversity is 1 regardless of --- this is the Swerling-0 (constant cross-section) case. The actual gain depends on the correlation structure between bistatic angles.
ex05-phased-mimo-tradeoff
MediumAn ISAC system has Tx antennas. The system uses phased-MIMO with independent waveforms.
(a) Express the Tx coherent gain per sub-array as a function of .
(b) Express the number of virtual elements as a function of (assuming ).
(c) Find that maximises the product (coherent gain) (virtual elements), representing a balanced ISAC metric.
Sub-array size is ; coherent gain is .
Virtual elements: .
Coherent gain
Each sub-array has elements. Coherent gain: .
Virtual elements
.
Optimal $L$
Metric: . This is monotonically decreasing in , so the metric favours small (more gain, fewer virtual elements).
A more balanced metric is --- constant! This shows the fundamental phased-MIMO trade-off: the product of beamwidth and number of beams is constant. The choice of depends on the specific ISAC requirements.
ex06-bistatic-range
MediumA bistatic radar has Tx at m and Rx at m (baseline m). A target is at position m.
(a) Compute the bistatic range .
(b) Compute the bistatic angle .
(c) If the bandwidth is MHz, what is the bistatic range resolution at this geometry?
where is the Euclidean distance.
The bistatic angle is the angle at the target in the Tx-target-Rx triangle.
Distances
m. m.
Bistatic range
m.
Bistatic angle
Using the law of cosines in the Tx-target-Rx triangle: . . . .
Range resolution
m.
ex07-kspace-samples
MediumA monostatic radar at the origin operates at two frequencies GHz and GHz with a 4-element ULA (, mm).
(a) How many k-space samples does this system produce per angular direction ?
(b) Sketch (or describe) the k-space coverage for a scene spanning .
(c) What is the range resolution provided by the two frequencies?
Monostatic: .
Two frequencies create two concentric arcs in k-space.
k-space samples per direction
With virtual elements (monostatic: Tx = Rx array) and 2 frequencies: total measurements, but all at the same angular direction. Per direction, there are 2 k-space samples (one per frequency) with different radial positions.
k-space coverage
At frequency : an arc at radius spanning with 16 samples. At frequency : a similar arc at slightly larger radius. The coverage is two concentric arcs.
Range resolution
Effective bandwidth: GHz. Range resolution: m.
ex08-condition-number
MediumA MIMO sensing matrix has Kronecker structure where and .
(a) What is ?
(b) If the Rx steering matrix has and the Tx steering matrix has , verify that is consistent.
(c) To improve to below 12, which factor should be improved first?
.
Part (a)
.
Part (b)
. Consistent.
Part (c)
Currently . Target: . Improving from to gives . Alternatively, improving from 3 to 1.5 gives . The largest factor () offers the most leverage.
ex09-coprime-array
HardA coprime MIMO array has Tx elements at positions and Rx elements at positions where .
(a) Compute all 15 virtual positions.
(b) How many unique virtual positions are there? Is the virtual array a filled ULA?
(c) Compare with the standard MIMO design (Tx at , Rx at ). Which has more unique virtual positions?
Virtual position: .
For the standard design, the virtual array is a filled ULA of elements.
Coprime virtual positions
All 15 sums : . (Some values appear only once; no duplicates in this case.) There are 15 unique positions out of 15 pairs.
Virtual array structure
The virtual positions span . This is not a filled ULA --- positions are missing. However, the aperture extends to , much larger than for the standard design.
Comparison
Standard design: filled ULA, 15 elements, aperture . Coprime design: 15 elements (no redundancy), aperture , but with 7 holes.
The coprime design achieves larger aperture (better angular resolution) at the cost of missing virtual positions (higher sidelobes, worse conditioning).
ex10-multiview-resolution
HardA multi-static MIMO system has terminals equally spaced on an arc of angular aperture .
(a) Derive the cross-range resolution as a function of .
(b) At GHz, find the minimum to achieve cm.
(c) If each terminal has a co-located MIMO array with , what is the per-terminal angular resolution? At what range does the multi-view resolution surpass the per-terminal resolution?
.
Per-terminal resolution depends on range: .
Cross-range resolution
From diffraction tomography: .
Minimum angular aperture
mm. Need mm: , so , , .
Crossover range
Per-terminal: , . . Multi-view (at ): mm. Crossover: mm, mm cm. For cm, multi-view is better.
ex11-kspace-filling
HardShow that for a monostatic system with equally spaced frequencies from to and virtual array elements spanning angle , the k-space coverage forms a polar-coordinate grid with:
(a) Radial extent .
(b) Angular extent .
(c) Derive the 2D PSF (point spread function) as the 2D sinc of the k-space aperture.
For monostatic, , so .
Radial extent
ranges from to . Extent: .
Angular extent
At fixed , the angular span is determined by . The transverse extent in k-space is .
2D PSF
The image is the inverse Fourier transform of the k-space coverage (indicator function). For a rectangular k-space aperture (linearised polar Cartesian):
The 3 dB widths are: down-range , cross-range .
ex12-kronecker-mvp
HardConsider a MIMO sensing matrix with Kronecker structure where (frequency range bins) and (virtual elements angle bins). The total scene has voxels.
(a) Show that the matrix-vector product can be computed as where is the reshaped scene matrix.
(b) What is the computational cost of this factored approach vs. direct multiplication?
(c) If both and are partial DFT matrices, how can FFTs further reduce the cost?
Reshape into .
For Kronecker products: .
Kronecker matrix-vector product
By the mixed-product property of Kronecker products: . So and .
Computational cost
Direct: . Factored: costs , then costs . Total: --- much less when .
FFT acceleration
If is a partial DFT (ULA steering vectors), is a column-wise FFT: . Similarly is a row-wise FFT: . Total: .
ex13-near-field
MediumA co-located MIMO array has aperture m and operates at GHz ( mm).
(a) Compute the far-field boundary .
(b) For a target at range m, is the far-field assumption valid? What about m?
(c) For a distributed MIMO system with baseline m at the same frequency, compute . Comment on the implications.
.
Co-located far-field
m. At m: near-field (need spherical wavefront model). At m: deep near-field.
Distributed far-field
m = 1300 km! Any practical scene is in the extreme near-field of the distributed array. Planar wavefront assumptions are invalid.
ex14-gram-matrix
MediumFor a 2-element Tx and 2-element Rx ULA at , with frequency, compute the Gram matrix for a scene with angles . Identify the pair with highest mutual coherence.
The virtual array has 4 elements at spacing.
Coherence between columns : .
Virtual array steering vectors
Virtual array: 4 elements at spacing . .
Gram matrix
where .
Highest coherence pair
Compute for all pairs. vs. : . vs. : . The pair has the smallest and hence the highest coherence --- these are the hardest to resolve.
ex15-sync-error
MediumA distributed MIMO system at GHz has a time synchronisation error of ns between two nodes.
(a) What range error does this introduce?
(b) What phase error does this correspond to at the carrier?
(c) If the system uses OFDM with kHz subcarrier spacing, how many subcarriers of phase drift does cause across the bandwidth?
Range error: .
Phase error: .
Range error
m.
Phase error
rad. This is 5 full cycles --- completely destroys phase coherence.
OFDM phase drift
Phase slope across frequency: . Phase change per subcarrier: rad . Across 1000 subcarriers: rad . The timing error manifests as a linear phase ramp across subcarriers, which can be estimated and removed.
ex16-optimal-partition
Challenge(Open-ended.) Consider the problem of partitioning physical antennas into Tx and Rx groups to minimise the condition number of the virtual array sensing matrix.
(a) For a ULA of elements at spacing, argue that minimises (assume filled virtual ULA).
(b) For a general (non-ULA) placement of elements, formulate the combinatorial optimisation problem.
(c) Propose a greedy algorithm that approximately solves this problem and analyse its complexity.
For ULAs, equal partition maximises both and minimises the condition number ratio.
The combinatorial problem has possible partitions.
ULA case
For a filled virtual ULA, (DFT matrix is unitary). The question reduces to maximising the virtual aperture length subject to . The aperture is , maximised at by AM-GM.
Combinatorial formulation
Given element positions , find partition minimising . This is NP-hard in general (related to minimum condition number column subset selection).
Greedy algorithm
Start with all elements as Rx (). Iteratively move the element to Tx that most reduces , stopping when . Complexity: condition number evaluations, each for small arrays. Total: --- feasible for .
ex17-nufft
ChallengeWhen the MIMO sensing matrix lacks Kronecker structure (distributed, near-field), the matrix-vector product costs . The non-uniform FFT (NUFFT) approximates this as:
(a) Explain why the k-space samples are non-uniformly spaced in the distributed case.
(b) The NUFFT has complexity . For what ratio does the NUFFT become faster than direct computation?
(c) Discuss the approximation error of the NUFFT and how it affects image reconstruction.
Non-uniform k-space arises because different Tx-Rx pairs have different bistatic angles.
NUFFT uses oversampled FFT + interpolation (Gaussian kernels or Kaiser-Bessel).
Non-uniform k-space
In the distributed case, depends on the direction to the target from each antenna. Since the antennas are at arbitrary positions, the resulting k-space samples are non-uniformly distributed --- unlike the co-located case where they lie on a regular grid.
Complexity crossover
Direct: . NUFFT: . NUFFT is faster when , i.e., , so . For : . The NUFFT is faster for essentially all practical configurations.
Approximation error
The NUFFT approximation error depends on the oversampling factor and the kernel width . Typical choices (, ) give relative error , which is negligible compared to measurement noise. However, the adjoint must use the same NUFFT approximation to preserve the self-adjoint structure needed for iterative algorithms.