E4702 HW#1 solutionsby Anmo Kim (ak2263@columbia.edu)

office hour: 2-3PM TUE&THR (1247 Mudd)

1. (P1.15) A running integrator is defined by

y(t) = t

tTx()d (1)

where x(t) is the input, y(t) is the output, and T is the integration period. Both x(t) and y(t) are samplefunctions of stationary processes X(t) and Y (t), respectively. Show that the power spectral density of theintegrator output is related to that of the integrator input by

SY (f) = T 2sinc2(fT )SX(f) (2)

(Solution)

The relation between SY (f) and SX(f) can be described by (1.58) in Haykins.

SY (f) = |H(f)|2SX(f)

Thus, we need to show that |H(f)|2 = T 2sinc2(fT ).The impulse response h(t) can be directly obtained from (1) by giving the impulse function, (t), as an input.

h(t) = { 1, 0 t T0, otherwise

By the Fourier transform,

H(f) = T

0

1 ej2ftdt

=1

j2f(ej2fT 1)

=1

2f [sin 2fT + j(cos 2fT 1)]

Then,

|H(f)|2 = 1(2f)2

[(sin 2fT )2 + (cos 2fT 1)2]

=1

(2f)2 [2 2 cos 2fT ]

=4 sin2 fT

(2f)2

= T 2sinc2(fT )

Thus,SY (f) = T 2sinc2(fT )SX(f)

1

2. (P1.16) A zero-mean stationary process X(t) is applied to a linear filter whose impulse response is defined bya truncated exponential:

h(t) = { aeat, 0 t T0, otherwise

Show that the power spectral density of the filter output Y (t) is

SY (f) =a2

a2 + 42f2(1 2 exp(aT ) cos(2fT ) + exp(2aT ))SX(f)

where SX(f) is the power spectral density of the filter input.

(Solution)

Since X(t) is a stationary process, then as in (P1.15), SY (f) can be described by the following equation.

SY (f) = |H(f)|2SX(f)

The Fourier transform of the given h(t) is

H(f) =

h(t)ej2ftdt

= T

0

ae(a+j2f)tdt

= aa + j2f

[e(a+j2f)T 1

]

Then,

|H(f)|2 = H(f)H(f)

={ a

a + j2f[e(a+j2f)T 1

]}{ a

a j2f [e(aj2f)T 1

]}

=a2

a2 + 42f2 [1 2eaT cos(2fT ) + e2aT ]

Thus,

SY (f) =a2

a2 + 42f2(1 2 exp(aT ) cos(2fT ) + exp(2aT ))SX(f)

3. (P1.17) The output of an oscillator is described by

X(t) = A cos(2Ft)where A is a constant, and F and are independent random variables. The probability density function of is defined by

f() = {12 , 0 20, otherwise

Find the power spectral density of X(t) in terms of the probability density function of the frequency F . Whathappens to this power spectral density when the frequency assumes a constant value?

(Solution)

Because and F are independent, the joint probability density function becomes:

f,F (, x) = f() fF (x)

2

Then, autocorrelation function of X(t) becomes:

RX(t, u) = E[X(t)X(u)]

=

x

f,F (, x)X(t)X(u)ddx

=

20

12

fF (x)X(t)X(u)ddx

=12

fF (x) 2

0

A2 cos(2xt ) cos(2xu )ddx

=A2

2

fF (x)12

20

[cos(2x(t + u) 2) + cos(2x(t u))] ddx

=A2

2

fF (x) cos(2x(t u))dx

Let = t u,RX() =

A2

2

fF (x) cos(2x)dx (3)

And RX() can be represented as a Fourier transform pair of power spectral density, SX(x):

RX() =

SX(x) exp(j2x)dx (4)

Since RX() is a real and even function,

RX() =

SX(x) cos(2x)dx (5)

From the similarity between (3) and (5), we can get SX(x) without calculating the Fourier transform ofRX() directly.

SX(x) =A2

2fF (x)

If the frequency x is a constant value, x0, fF (x) = 12(x x0) + 12(x + x0). Thus, SX(x) becomes:

SX(x) =A2

4(x x0) + A

2

4(x + x0)

3