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**EECS 20. Midterm No. 2 Solutions
November 8, 2002.
**

1. **40 points (10 points each part). **Consider the Simulink diagram shown below:

1

Out1

1 s

Integrator1

1 s

Integrator

.9

Gain

1

In1

*yx
*

This shows an LTI system with one input and one output, both of which are continuous-time
signals. The input and output are indicated by the rounded boxes, and are labeled *x *and *y*.
The gain is 0.9, and the integrators both have initial condition equal to 0.

(a) Write a differential equation (with no integrals, just derivatives) that relates the input *x
*and the output *y*.
**Answer**:

*∀ t ∈ Reals*+*, *0*.*9*x*(*t*) = *ÿ(t).
*(b) Give the [*A, b, c, d*] representation of this system.

**Solution**: Let the state be the outputs of the two integrators, as follows:

*∀ t ∈ Reals*+*, z*(*t*) =
[
*y*(*t*)
*ẏ(t)
*

]

Hence,

*A *=

[ 0 1 0 0

]
*b *=

[
0
0*.*9

]
*cT *= [1 0] *d *= 0*.
*

(c) Find the frequency response *H*:*Reals → Reals *of this system.
**Solution**: Let the input be *x*(*t*) = *eiωt*, in which case the output must be *H*(*ω*)*eiωt*.
Substituting these into the differential equation from part (a), we get

0*.*9*eiωt *= *H*(*ω*)(*−ω*2*eiωt*)*,
*which we can solve for *H*(*ω*) to get

*H*(*ω*) = *−*0*.*9*/ω*2*.
*(d) Find the output of the system if the input *x *is given by

*∀ t ∈ Reals, x*(*t*) = cos(2*t*)*.
***Solution**: Write this input as

*∀ t ∈ Reals, x*(*t*) = 0*.*5(*ei*2*t *+ *e−i*2*t*)*,
*which using linearity implies that the output must be

*∀ t ∈ Reals, y*(*t*) = 0*.*5(*H*(2)*ei*2*t *+*H*(*−*2)*e−i*2*t*) = 0*.*5((0*.*9*/*4)*ei*2*t *+ (0*.*9*/*4)*e−i*2*t*) = 0*.*2 cos(2*t*)*.
*

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2. **30 points (5 points each part). **Consider continuous-time systems with input *x*:*Reals →
Reals *and output *y*:*Reals → Reals*. Each of the following defines such a system. For each,
indicate whether it is linear (L), time-invariant (TI), both (LTI), or neither (N). Note that no
partial credit will be given for these questions.

(a) *∀ t ∈ Reals, ẏ(t) *= *x*(*t*) + 0*.*9*y*(*t*) **Answer**: LTI
(b) *∀ t ∈ Reals, y*(*t*) = cos(2*πt*)*x*(*t*) **Answer**: L
(c) *∀ t ∈ Reals, y*(*t*) = *x*(*t− *1) **Answer**: LTI
(d) *∀ t ∈ Reals, y*(*t*) = *x*(*t*) + 0*.*1(*x*(*t*))2 **Answer**: TI
(e) *∀ t ∈ Reals, y*(*t*) = *x*(*t*) + 0*.*1(*x*(*t − *1))2 **Answer**: TI
(f) *∀ t ∈ Reals, y*(*t*) = 0 **Answer**: LTI

3. **40 points (10 points each part). **Consider a discrete-time signal *x*: *Integers → Reals *defined
by

*∀ n ∈ Integers, x*(*n*) = 1*− *cos(3*πn/*4)*.
*

Assume this signal is sampled at 8,000 samples/second.

(a) Give the frequency of the cosine term in Hz (cycles/second).

**Solution**: The frequency is 3*π/*4 radians/sample. Divide by 2*π *radians/cycle and mul-
tiply by 8000 samples/second to get 3000 Hz.

(b) Give period of *x*.

**Solution**: The period is the smallest positive integer *p *such that 3*πp/*4 is a multiple of
2*π*. Thus, *p *= 8.

(c) Give the fundamental frequency (in any units, but be sure to give the units).

**Solution**: *ω*0 = 2*π/p *= *π/*4 radians/sample. Alternatives: 1/8 cycles/sample, or 8000/8
cycles/second = 1000 cycles/second.

(d) Give the coefficients *A*0*, A*1*, A*2*, · · · , AK *and *φ*1*, φ*2*, · · · , φK *of the Fourier series ex-
pansion for *x*,

*x*(*n*) = *A*0 +
*K*∑
*k*=1

*Ak *cos(*kω*0*n*+ *φk*)

where

*K *=

{
(*p− *1)*/*2 if *p *is odd
*p/*2 if *p *is even

**Solution**: *A*0 = 1, *A*1 = 0, *A*2 = 0, *A*3 = 1, *A*4 = 0, and *φ*1 = 0, *φ*2 = 0, *φ*3 = *π*,
*φ*4 = 0, although the phases corresponding to zero amplitude can be anything. The
*φ*3 = *π *accounts for the minus sign in the definition of *x*.

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