## Search in the document preview

**EECS 120. Solutions to Midterm No. 1, February 17, 2000.
**

1. **20 points **Find the expression for the frequency response from *x *to *y *in terms of
*H*1*,H*2*,H*3 for the system depicted in:

(a) Part (a) of Figure 1.
**Ans **Fix *ω ∈ Reals *and take as input the signal *x *: *t → X*(*ω*)*ejωt*. Since *H*1
is LTI, the other signals are of the form *w *: *t → W *(*ω*)*ejωt *and *y *:*→ Y *(*ω*)*ejωt*.
Moreover,

*W *(*ω*) = *X*(*ω*) + *Y *(*ω*)*, Y *(*ω*) = *H*1(*ω*)*W *(*ω*)*,
*

from which we obtain the desired frequency response,

*H*4(*ω*) =
*Y *(*ω*)
*X*(*ω*) =

*H*1(*ω*)
1*−H*1(*ω*)

(b) Part (b) of Figure 1.
**Ans **Rewrite part (b) of the figure as (c), and introduce the signal names *u, v*.
Take as input the signal *x *: *t → X*(*ω*)*ejωt*. Since *H*1*,H*2*,H*3 are all LTI, *u, v, y
*are all exponential,

*u *: *t → U*(*ω*)*ejωt, v *: *t → V *(*ω*)*ejωt, y *: *t → Y *(*ω*)*ejωt.
*From the previous part we know *V *(*ω*) = *H*4(*ω*)*U*(*ω*). And then, following the
same argument as in the previous part,

*Y *(*ω*)
*X*(*ω*)

=
*H*4(*ω*)*H*3(*ω*)

1 *−H*2(*ω*)*H*3(*ω*)*H*4(*ω*) *.
*

And substituting for *H*4 from the previous part,

*Y *(*ω*)
*X*(*ω*) =

*H*4(*ω*)*H*3(*ω*)
1*−H*2(*ω*)*H*3(*ω*)*H*4(*ω*) =

*H*1(*ω*)*H*3(*ω*)
1*−H*1(*ω*)*−H*1(*ω*)*H*2(*ω*)*H*3(*ω*)

2. **20 points **Let *f, g, x, y *be as in Figure 2.

(a) Determine *f ∗ g*.
**Ans **Since *f *is periodic with period 2, *f ∗ g *is periodic with period 2 and it can
be obtained as the periodic repetition of *f*1 *∗ g *where *f*1(*t*) = *f*(*t*)*, *0 *≤ t ≤ *2 and
*f*1(*t*) = 0, otherwise. *f*1 and *f*1 *∗ g *are shown graphically in the figure.

(b) Determine *x ∗ y*.
**Ans **Arguing in the same way, *x ∗ y *is the periodic repetition of *x*1 *∗ y*, with
period 1.5, and we can use the calculation of *f*1 *∗ g *as shown in the figure.

3. **20 points **Give an example of a discrete-time system *H *that is:

(a) Not linear;
**Ans **Take *H*(*x*)(*n*) = (*x*(*n*))2. This is not linear since *H*(2*x*) = 4*H*(*x*) = 2*H*(*x*).

1

(a)

*x y
*

*H2*ω)

*H3*(ω)+

(b)

*x H1*(ω)+ *y
*

*H1*(ω)

*x y
*

*H2*ω)

*H3*(ω)+

*(c)
*

*H1*(ω)+

*w
*

*H4*(ω)

*u v
*

*H4*(ω)

Figure 1: System for Problem 1

1

1 2-1-2

*f
*

1

0.5

*g
*

1.5-1 0.5

1

1

*x
*

*y
*

1

-1.5

20

*f1
*

20

0.5

0.5 1.0

*f1***g
*

0 1.5

*y1
*

1.50

0.5

0.5 1.0

*y1***x
*

Figure 2: Signals for Problem 2

2

(b) Linear and time-varying;
**Ans **Take *H*(*x*)(*n*) = *x*(2*n*). This is not time-invariant, since *H*(*D*1*x*)(*n*) =
*x*(2*n − *1) = *D*1(*Hx*)(*n*) = *H*(*x*)(*n− *1) = *x*(2*n − *2).

(c) LTI but not causal;
**Ans **Take *H*(*x*)(*n*) = *x*(*n*+ 1). This is not causal, because its impulse response
is *n → δ*(*n *+ 1), so that *h*(*−*1) = 1 = 0.

(d) LTI, causal, but not memoryless.
**Ans **Take *H*(*x*)(*n*) = *x*(*n− *1). This is not memoryless, because the output at
*n *depends on the input at *n− *1.

4. **20 points **Suppose a periodic signal *x *: *Reals → Comps *with fundamental frequency
*ωx *has the Fourier series representation:

*∀t, x*(*t*) =
*∞*∑

*k*=*−∞
Xke
*

*jkωxt.
*

(a) Let *y *be the signal *∀t, y*(*t*) = *x*(*t− τ*), where *τ *is a fixed number. What is the
Fourier series representation of *y*?
**Ans **We have *∀t*,

*y*(*t*) = *x*(*t− τ*) =
*∞*∑

*k*=*−∞
Xke
*

*jkωx*(*t−τ*) =
*∞*∑

*k*=*−∞
Xke
*

*−jωxτejkωxt
*

=
*∞*∑

*k*=*−∞
Yke
*

*jkωyt,
*

where

*Yk *= *Xke−jωxτ *and *ωy *= *ωx.
*

(b) Let *z *be the signal *∀t, z*(*t*) = *x*(2*t*). What is the fundamental frequency *ωz *of *z
*in terms of *ωx*? What is the Fourier series representation of *z*?
**Ans **We have *∀t*,

*z*(*t*) = *x*(2*t*) =
*∞*∑

*k*=*−∞
Xke
*

*jk*2*ωxt
*

*∞*∑

*k*=*−∞
Zke
*

*jkωzt
*

where

*ωz *= 2*ωx *and *Zk *= *Xk.
*

(c) Let *w *be the signal *∀t, w*(*t*) = *z*(*−t*). What is the Fourier series representation
of *w*?

3

**Ans **We have *∀t*,

*w*(*t*) = *z*(*−t*) =
*∞*∑

*k*=*−∞
Zke
*

*−jkωzt
*

=
*∞*∑

*k*=*−∞
Wke
*

*jkωwt
*

where

*ωw *= *ωz *= 2*ωx *and *Wk *= *Z−k *= *X−k.
*

4