Exercises
ex-ris-ch02-01
EasyHow many PIN diodes per element are needed to realize a -bit RIS? Answer for .
Each PIN diode has two states; adding diodes gives combinations.
Formula
bits require PIN diodes (each toggling a different fraction of the cell's impedance to yield distinct phase states).
Cases
1-bit: 1 diode, 2 phase states. 2-bit: 2 diodes, 4 states. 3-bit: 3 diodes, 8 states. Note: this is the straightforward binary encoding; alternative designs (varactor + DAC) use a single tunable element.
ex-ris-ch02-02
EasyCompute the exact -bit quantization SNR loss (in dB) for .
.
Plug in
. . . . .
Interpretation
The return on additional bits diminishes sharply after 3 bits. Going from 2 → 3 bits saves ; 4 → 5 bits saves only — not worth the doubled diode count.
ex-ris-ch02-03
MediumProve that the sinc approximation of the quantization loss assumes the quantization errors are uniformly distributed on . Under what condition on the optimal phases does this assumption hold?
Recall the 'dithering' condition for uniform-rounding noise.
If are clustered at particular values, the quantization error is not uniform.
Uniform-error condition
The error is uniform on iff is uniform on . This holds, e.g., if the optimal phases themselves are i.i.d. uniform on — which is the case when the channels have uniform random phases (generic Rayleigh).
When it fails
In pure LoS scenarios with a regular array geometry, the optimal phases follow a quadratic spatial pattern (steering vector), and their values-mod- may cluster unfavorably. In this case the quantization loss can be slightly better or worse than the sinc prediction; on average the two bracket agree.
ex-ris-ch02-04
MediumA unit cell has amplitude profile . Compute the average amplitude over and the corresponding APC loss in dB.
Use .
Compute mean
.
APC loss
, loss .
ex-ris-ch02-05
MediumAn RIS has elements, 2-bit phase control, and unit-cell APC with . Estimate the combined hardware SNR penalty.
Combine quantization loss and APC loss multiplicatively.
Individual losses
Quantization (): , . APC: , .
Combined
Total SNR factor , . Or additively in dB: .
ex-ris-ch02-06
EasyDescribe in one sentence when the diagonal RIS model breaks down most severely.
Think about element spacing and coupling.
Answer
When element spacing is substantially smaller than half-wavelength (e.g., or less), mutual coupling becomes strong, off-diagonal terms in cannot be ignored, and beam-pattern predictions from the diagonal model can be off in main-lobe gain.
ex-ris-ch02-07
HardSuppose a 1-bit RIS must achieve the same SNR as a continuous-phase RIS of size . How large must the 1-bit RIS be?
Write the two SNRs as and .
Set SNRs equal
Continuous: . 1-bit: . Equal: , so .
Interpretation
The 1-bit RIS needs about more elements to match a continuous-phase RIS. A -continuous becomes 1-bit. Cost-wise, 400 1-bit elements (each 1 diode) vs. 256 continuous elements (each DAC + varactor) can go either way — depending on whether the DAC or the extra elements are cheaper.
ex-ris-ch02-08
MediumWhy does the reflection phase of a grounded LC resonator span approximately as the reactance passes through zero?
Look at .
Phase formula
For purely imaginary , . . Simplifying, (modulo branch choices; in practice one sweeps and tracks ).
Limit behaviour
As (capacitive-dominated, below resonance), . As (inductive, above resonance), . But passing through makes jump by , so sweeps continuously from down through and back to — covering total.
Interpretation
The resonance provides the only way to achieve a full phase range with a single reactive element. Off-resonance circuits (no inductor/capacitor cancellation) span only or less of phase.
ex-ris-ch02-09
MediumAn RIS has elements and 3-bit phase control. If each element's phase is updated at , compute the required control-link bandwidth.
Bits per update × updates per second.
Bits per update
bits per full reconfiguration.
Bandwidth
bps kbps. Easily carried by even the cheapest out-of-band control link (Bluetooth LE, Zigbee, or a wired serial bus). Control overhead is not a bottleneck for low-rate update. At kHz update rates (Chapter 18) the bandwidth becomes Mbps, which starts to demand a dedicated link.
ex-ris-ch02-10
ChallengeConsider a tridiagonal mutual-impedance model from Example 2.4. For a 1D array of elements, derive an expression for the eigenvalues of the coupling matrix in the limit .
Tridiagonal Toeplitz matrices have eigenvalues given by a closed form: .
Toeplitz formula
For a symmetric tridiagonal Toeplitz matrix with diagonal and off-diagonal , eigenvalues are , .
Apply to RIS
With (from diagonal) and (if we take the magnitude for a symmetric model), eigenvalues spread around by .
Infinite limit
As , the eigenvalues densely fill the interval . The diagonal approximation replaces this -spread distribution with a single value; the inversion error in is of order in the worst case. This motivates either explicit coupling modeling or design-for-isolation (spacing choices).
ex-ris-ch02-11
MediumAn RIS element has reflection phase error approximately per element due to manufacturing variance. Estimate the coherent-SNR loss for a -element RIS.
Random phase errors reduce the coherent sum by for small Gaussian errors.
Expected field reduction
For Gaussian (in radians), . , so .
Power loss
Power is amplitude squared: , . Negligible.
Comparison with quantization
Manufacturing variance of is much smaller than the step of a 4-bit quantizer. The loss is accordingly smaller. Manufacturing tolerances at the few-degree level are not a first-order concern for modern designs.
ex-ris-ch02-12
HardSuppose the unit-modulus assumption fails in the following structured way: for and for (a half-circle reflector). Assuming optimal phases are uniformly distributed on , compute the effective APC penalty.
average of 1 and 0.5.
Average amplitude
Half the land in with and half in with . .
APC loss
, .
Design implication
Half-circle reflectors are cheap (one diode, one reactance switch) but cost the equivalent of a 1-bit quantizer on top of the amplitude loss — around - total.
ex-ris-ch02-13
MediumWhy is it physically impossible for a fully passive RIS element to simultaneously achieve and arbitrary ?
Passivity requires .
Resonance-induced phase rotation always involves some ohmic loss in the diode.
Passivity bound
Passivity dictates ; equality requires a perfectly lossless reflection.
Resonance and loss
To rotate the phase by up to with a single tunable element, the element must pass through resonance. Near resonance, the small (but non-zero) loss resistance of any real diode causes to dip below 1. The only way to achieve exact unit modulus across all phases is to add an active amplifier — but then the element is no longer passive.
Conclusion
across all is a strictly unreachable design target for passive hardware. Real designs accept a few-dB amplitude dip near resonance as the price of reconfigurability.