ECE251D: Homework #5 Solutions - UCSD DSP LAB | The ...

ECE251D: Homework #5 Solutions

Problem 4.2.1

Consider an 8-element circular array with equal spacing between elements. It could also be viewedas two 4-element rectangular arrays with the first array oriented along the x-y axes and the secondarray rotated by 45◦.

(a)Using this model, find the beam pattern and compare your result to the result in the text.

The distance between two elements on the same axis is d. Then, the elements for one rectangulararray are at (d/2, 0, 0), (−d/2, 0, 0), (0, d/2, 0) and (0,−d/2, 0). The weights are 1/8 each. So thebeampattern is given by

BR1(θ, φ) =1

8

[ejk0

d2sin θ cosφ + e−jk0

d2sin θ cosφ + ejk0

d2sin θ sinφ + e−jk0

d2sin θ sinφ

]=

1

4cos

(k0

d

2sin θ cosφ

)+

1

4sin

(k0

d

2sin θ sinφ

)where k0 =

2πλ .

Similarly, for the other array, rotated by 45◦ around the x-y axis, the beam pattern is given by

BR2(θ, φ) = BR1(θ, φ− π

4) =

1

4cos

(k0

d

2sin θ cos

(φ− π

4

))+

1

4sin

(k0

d

2sin θ sin

(φ− π

4

))

Thus, the beam patern is given by

B(θ, ϕ) = (BR1(θ, φ) +BR2(θ, φ))

=1

4

[cos

(k0

d

2sin θ cosφ

)+ sin

(k0

d

2sin θ sinφ

)+cos

(k0

d

2sin θ cos

(φ− π

4

))+ sin

(k0

d

2sin θ sin

(φ− π

4

))](1)

The beam pattern of circular array with N = 8 is given by (eq. 4.171)

B(θ, φ) =N−1∑m=0

Bm(θ, φ)

1

whereBm(θ, φ) = wmejkR sin θ cos(φ−φm)+jβn

βn = −kR sin θ0 cos(φ− φn)

Setting wm = 1/N = 1/8, βn = 0 since θ0 = 0, φm = 2πmN , and R = d/2, the expression for beam

pattern is

Bm(θ, φ) =1

8ejk0

d2sin θ cos(φ− 2πm

8)

and

B(θ, φ) =

7∑n=0

Bm(θ, φ) =

3∑m=0

B2m(θ, φ) +

3∑m=0

B2m+1(θ, φ)

Examining the first term:

3∑m=0

B2m(θ, φ) =1

8

3∑m=0

ejk0d2sin θ cos(φ−π

2m)

=1

8

[ejk0


d2sin θ cos(φ− π

L) + ejk0

d2sin θ cos(φ−π) + ejk0

d2sin θ cos(φ− 3π

2)]

=1

8

[ejk0


d2sin θ sinφ + e−jk0

d2sin θ cosφ + e−jk0

d2sin θ sinφ

]=

1

4

[cos(k0

d

2sin θ cosφ) + cos(k0

d

2sin θ sinφ)

]= BR1(θ, φ)

Similarly,3∑

m=0

B2m+1(θ, φ) = BR2(θ, φ)

∴ B(θ, φ) = BR1(θ, φ) +BR2(θ, φ)

The same as Equation 1.

(b) Assume d = λ in the rectangular arrays. Plot the resulting beam pattern.Substituting d = λ in Equation 1, we have

B(θ, ϕ) =1

4

[cos(π sin θ cosφ) + sin(π sin θ sinφ) + cos

(π sin θ cos

(φ− π

4

))+ sin

(π sin θ sin

(φ− π

4

))]

We plot the beam pattern cuts of this function in Figure 1. The matlab code is attached inAppendix A. We have normalized the beam pattern thus the peak value is 0 dB, not near −10 dB.

2

−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−50

−45

−40

−35

−30

−25

−20

−15

−10

−5

0

5

ur

Bea

m p

atte

rn (

dB)

Circular Array Beam pattern Cut

φ = 0°

φ = 45°

φ=90°

Figure 1: Beam pattern cuts for a uniformly weighted circular array (N=8) for different values ofφ.

A Matlab Code for Prob. 4.2.1

% File: t421b.m

% Created: Wed 11/03/04 03:13:12 -0800

% Last Updated: Thu 11/04/04 03:13:20 -0800

% Description: ECE 251D - Fa04 - HW5 - 4.2.1b van Trees

% : Beam Pattern Cuts - Circular Array

% : N=8, d = lambda

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% We have developed upon the files provided by Prof. Bell (GMU) at

% http://ite.gmu.edu/DetectionandEstimationTheory/

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%

close all

clear all

N = 8;

ur = -1:1/1000:1;

theta = pi/2 * ur;

sintheta = sin(theta);

PHI_SET = [0 pi/4 pi/2];

B = zeros(length(PHI_SET), length(ur));

for k=1:length(PHI_SET)

phi = PHI_SET(k);

B(k,:) = B(k,:) + cos(pi * cos(phi) * sintheta);

B(k,:) = B(k,:) + sin(pi * sin(phi) * sintheta);

B(k,:) = B(k,:) + cos(pi * cos(phi-pi/4) * sintheta);

B(k,:) = B(k,:) + sin(pi * sin(phi-pi/4) * sintheta);

end

B = B./max(max(abs(B)));

figure;

hold off

plot(ur, 10 * log10(abs(B(1,:))),’b-’);

3

hold on

plot(ur, 10 * log10(abs(B(2,:))),’g--’);

plot(ur, 10 * log10(abs(B(3,:))),’r-.’);

axis([-1 1 -50 5]);

xlabel(’\it u_r’,’Fontsize’,14);

ylabel(’Beam pattern (dB)’,’Fontsize’,14);

title(’Circular Array Beam pattern Cut’);

h=legend(’\phi = 0^\circ’, ’\phi = 45^\circ’, ’\phi=90^\circ’, 3);

set(h,’Fontsize’,12)

4

ECE251D: Homework #6 Solutions

Problem 6.2.4

(a) The array gain for an MVDR beamformer is given by (6.28)

Ao = V H(ks)ρ−1n V (ks) = 1Hρ−1

n 1 (1)

where 1 is a vector of all ones and

ρn =[

σ2w

σ2w + M1

I +M1

σ2w + M1

VIVHI

]=

[1

1 + INRI +

INR

1 + INRVIV

HI

]

0 0.5 1 1.5 2 2.50

5

10

15

20

25

30

35

40

45Array Gain

uI / BW

NN

Ao (

dB)

INR=0 dBINR=10 dBINR=20 dBINR=30 dB

Figure 1: Array gain versus uI/BWNN for an MVDR beamformer, when INR = 0,10, 20 and 30dB.

Clearly the array gain depends on the INR. Figure 1 shows the array gain versus uI/BWNN for anMVDR beamformer, when INR = 0,10, 20 and 30 dB. It can be seen that the array gain increasesas the direction of the interference signal moves from the center of the mainlobe towards the endfire

1

angle. Once uI/BWNN becomes larger or equal to 0.5, which means the interference direction isoutside the mainlobe, the array gain tends to converge to a constant.

(b) The sensitivity ||wmvdr||2 is computed and shown in Figure 2. The larger condition number ofSn at higher INR, leads to a larger norm for the MVDR weights.

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

−10

−5

0

5

10

15

20

||Wmvdr

||2

uI / BW

NN

||Wm

vdr||2 (

dB)

INR=0 dBINR=10 dBINR=20 dBINR=30 dB

Figure 2: ||wmvdr||2 versus uI/BWNN for an MVDR beamformer, when INR = 0,10, 20 and 30dB.

Problem 6.2.8

(a) Note: The normalization in the book is incorrect, it should be

wdq =wdq

||wdq||

Problem: minWwHSnw subject to wHdqw = 1

This is the same problem as MVDR, where wdq replaces V (ks). The solution can readily beobtained by replacing V (ks) in the MVDR solution with wdq.

wmvqr = ΛqS−1n wdq

when Λq = 1

wHdqS−1

n wdq

.

2

(b)

SNR at input = SNRI =σ2

s

σ2n

SNR at output = SNRO =σ2

s |wHmvqrV (ks)|2

wHmvqrSnwmvqr

=σ2

s

σ2n

|wHmvqrV (ks)|2

wHmvqrρnwmvqr

∴ ArrayGain =SNRO

SNRI

=|wH

mvqrV (ks)|2wH

mvqrρnwmvqr

(c) Sn = σ2W I

Array gain =|wH

mvqrV (ks)|2||wmvqr||22

=|wH

dqS−1n V (ks)|2

wHdqS

−2n wdq

=|wH

dqV (ks)|2wH

dqwdq

= |wHdqV (ks)|2

Problem 6.3.3

The array gain for MVDR beamformer can be computed by Equation 1 (Prob 6.2.4).

The weight vector for the Kaiser beamformer with a null at uI is

wk = a · (I − C(CHC)−1CH) ·wkaiser

where C = V (uI) =

e−j N−12

π·uI

...ej N−1

2π·uI

, wkaiser is the Kaiser window and a is the normalizing

coefficient, so that the responses in the desired signal’s direction is distortionless.

wHK =

wHkaiser · (I − C(CHC)−1CH)

wHkaiser · (I − C(CHC)−1CH) · Vs

3

where wkaiser = I0

[β

√1− (

2nN

)2]

and n = −N−12 , ..., N−1

2

The beampattern is

Bk(u) = wHK · V (u) =

wHkaiser · (I − C(CHC)−1CH)

wHkaiser · (I − C(CHC)−1CH) · Vs

· V (u)

The array gain is computed as a ratio of the SNR at the output of the array to the ratio of theSNR at the input of a signal sensor.

Akaiser =SNRO

SNRI=

|wHK ·vS |2·σ2

s

wHK ·(M1·VIV H

I +σ2wI)·wK

σ2s

sI+σ2w

=(σ2

I + 1)wH

K · (σ2I · VIV H

I + 1) ·wK(2)

where we assumed distortionless response along the desired signal’s direction, Vs. (|wHK · Vs|2 = 1),

and INR= σ2I = M1

σ2w

.

In fact, Equation 2 can be used for any weighting scheme besides Kaiser weighting.

Table 1, 2 and 3 show the array gains of MVDR beamformer and Kaiser weighting beamformerfor interferences with different arrival directions and power. It can be seen that the array gains ofKaiser weighting beamformer are always less than those of MVDR beamformer. MVDR is betterable to maximize the SNR of the array output, thus the improved array gain.

The beam patterns of MVDR beamformer and Kaiser weighting beamformer are shown in thefollowing pages. MVDR beamformer has narrower mainlobe than Kaiser weighting beamformer,which means MVDR has better directivity.

When the interference comes from a direction outside the desired null-to-null bandwidth, i.e. |uI | ≥BWNN/2 = 0.2, the mainlobe is well shaped and pointed towards broadside. However, if theinterference direction gets into the null-to-null bandwidth interval, the mainlobe is significantlydistorted.

Table 1: Array gains for uI = 0.3 and 0.5, σ2I = 10dB and 50 dB.

uI = 0.3 uI = 0.5 uI = 0.3 uI = 0.5σ2

I = 10dB σ2I = 10dB σ2

I = 50dB σ2I = 50dB

MVDR 20.2001 dB 20.3271 dB 59.7840 dB 59.9123 dBKaiser 18.9183 dB 18.3077 dB 58.5044 dB 57.8938 dB

4


uI = 0.09 uI = 0.18 uI = 0.09 uI = 0.18σ2

I = 10dB σ2I = 10dB σ2

I = 50dB σ2I = 50dB

MVDR 17.5193 dB 20.3609 dB 57.0641 dB 59.9464 dBKaiser 12.6956 dB 18.8725 dB 52.2817 dB 58.4586 dB


uI = 0.02 uI = 0.0433 uI = 0.2 uI = 0.0433σ2

I = 10dB σ2I = 10dB σ2

I = 50dB σ2I = 50dB

MVDR 6.6187 dB 12.2369 dB 45.0720 dB 51.5740 dBKaiser -10.8994 dB 1.9869 dB 28.6867 dB 41.5730 dB

−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−140

−120

−100

−80

−60

−40

−20

0

20MVDR beam patterns

u

|B(u

)| (

db)

uI =0.3, σ

I2 =10dB

uI =0.5, σ

I2 =10dB

uI =0.3, σ

I2 =50dB

uI =0.5, σ

I2 =50dB

Figure 3: MVDR beamformer for uI = 0.3 and 0.5, σ2I = 10dB and 50 dB.

5

−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−350

−300

−250

−200

−150

−100

−50

0

50Kaiser beam patterns

u

|B(u

)| (

db)

uI =0.3, σ

I2 =10dB

uI =0.5, σ

I2 =10dB

uI =0.3, σ

I2 =50dB

uI =0.5, σ

I2 =50dB

Figure 4: Kaiser beamformer for uI = 0.3 and 0.5, σ2I = 10dB and 50 dB.

−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−140

−120

−100

−80

−60

−40

−20

0


u

|B(u

)| (

db)

uI =0.09, σ

I2 =10dB

uI =0.18, σ

I2 =10dB

uI =0.09, σ

I2 =50dB

uI =0.18, σ

I2 =50dB


6

−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−350

−300

−250

−200

−150

−100

−50

0


u

|B(u

)| (

db)

uI =0.09, σ

I2 =10dB

uI =0.18, σ

I2 =10dB

uI =0.09, σ

I2 =50dB

uI =0.18, σ

I2 =50dB


−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−100

−80

−60

−40

−20

0


u

|B(u

)| (

db)

uI =0.02, σ

I2 =10dB

uI =0.0433, σ

I2 =10dB

uI =0.02, σ

I2 =50dB

uI =0.0433, σ

I2 =50dB


7

−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−300

−250

−200

−150

−100

−50

0


u

|B(u

)| (

db)

uI =0.02, σ

I2 =10dB

uI =0.0433, σ

I2 =10dB

uI =0.02, σ

I2 =50dB

uI =0.0433, σ

I2 =50dB


8

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