Exam I: Solutions

BME 171-02, Signals and Systems

Exam I: Solutions

100 points total

1. (15 pts.) Fill in the following table (for each column, give a “yes/no” answer and brieflyjustify it):

System Linear Time-invariant Causal

y(t) = 3x(t) cos(t) yes no yes

y(t) =√

x2(t) no yes yes

y[n] =

{

+1, x[n] ≥ 0−1, x[n] < 0

no yes yes

y(t) =

∫ t+1

tx(λ)dλ yes yes no

y[n] = 2(x[n + 1]u[n] − x[n]) + 1 no no no

Exam I: Solutions

2. (15 pts.) Consider the following system:

Fx[n] y[n]

Δ

2 _

+ z[n]

Here, the system F is defined by the input-output relationship

F{

z[n]}

= z[n] − z[n − 1],

and ∆ is the unit delay

∆{

w[n]}

= w[n − 1].

Write down the linear difference equation describing this system.

Solution. Let z[n] be the output of the summer, as shown above. Then

y[n] = F{

z[n]}

= z[n] − z[n − 1].

Now,

z[n] = 2x[n] − ∆{

y[n]}

= 2x[n] − y[n − 1].

Therefore, substituting the expression for z[n] into the first equation, we can write

y[n] = z[n] − z[n − 1]

=(

2x[n] − y[n − 1]︸ ︷︷ ︸

=z[n]

)

−

(

2x[n − 1] − y[n − 2]︸ ︷︷ ︸

=z[n−1]

)

= 2x[n] − y[n − 1] − 2x[n − 1] + y[n − 2].

Simplify to get

y[n] + y[n − 1] − y[n − 2] = 2x[n] − 2x[n − 1]

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Exam I: Solutions

3. (20 pts.) Consider the following circuit:

+

_

x(t)

+

_

y(t)

R1 L

C R2

i0(t)

i1(t)

i2(t)

Write down the input-output differential equation for this circuit in terms of the input voltagex(t) and the output voltage y(t).

Solution. Let

i0(t) = current through R1

i1(t) = current through C

i2(t) = current through R2

Then, using Kirchhoff’s voltage law, we write

−x(t) + R1i0(t) + Ldi0(t)

dt+ y(t) = 0.

On the other hand, using Kirchhoff’s current law, we have

i0(t) = i1(t) + i2(t).

Moreover,

i1(t) = Cdy(t)

dtand i2(t) =

y(t)

R2.

Therefore,

i0(t) = Cdy(t)

dt+

y(t)

R2

anddi0(t)

dt= C

d2y(t)

dt2+

1

R2

dy(t)

dt.

Substituting this into the KVL equation, we get

−x(t) + R1

(

Cdy(t)

dt+

y(t)

R2

)

+ L

(

Cd2y(t)

dt2+

1

R2

dy(t)

dt

)

+ y(t) = 0.

Simplify to get the final answer:

LCd2y(t)

dt+(

R1C + LR2

)dy(t)

dt+(

R1

R2+ 1

)

y(t) = x(t).

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Exam I: Solutions

4. (25 pts.) Compute the convolution of the following two signals and plot the result on the setof axes provided. Show all your work!

x[n] ν[n]

n n0 0

1 1

-1

x[n] * ν[n]

n

1

0

Solution. First, rename the time variable n into k. Next, flip and shift one of the signals.We will flip ν[k] to get ν[−k]:

x[k] ν[-k]

k k0 0

1 1

-1

Now, shift by n to get ν[n − k]. Note that there is no overlap between x[k] and ν[n − k] aslong as n ≤ −2 or n ≥ 4:

x[k] ν[-2-k]

k k0 0

1 1

-1

x[k] ν[4-k]

k k0 0

1 1

-1

4

Name: ..................................................

So, y[n] = x[n] ⋆ ν[n] = 0 for n ≤ −2 and for n ≥ 4

Now, shift and compute the overlap:

y[−1] = x[−1]ν[0] + x[0]ν[−1] + x[1]ν[−2]

= (−1) · 1 + 0 · 0 + 1 · 0

= −1

y[0] = x[−1]ν[1] + x[0]ν[0] + x[1]ν[−1]

= (−1) · 1 + 0 · 1 + 1 · 0

= −1

y[1] = x[−1]ν[2] + x[0]ν[1] + x[1]ν[0]

= (−1) · 1 + 0 · 1 + 1 · 1

= 0

y[2] = x[−1]ν[3] + x[0]ν[2] + x[1]ν[1]

= (−1) · 0 + 0 · 1 + 1 · 1

= 1

y[3] = x[−1]ν[4] + x[0]ν[3] + x[1]ν[2]

= (−1) · 0 + 0 · 0 + 1 · 1

= 1

Overall,

y[n] = x[n] ⋆ ν[n] =

0, n ≤ −2−1, n = −1, 00, n = 11, n = 2, 30, n ≥ 4

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Exam I: Solutions

5. (25 pts.) Compute the convolution of the following two signals. Write down its analyticalform and sketch its plot in the set of axes provided. Show all your work!

x(t) ν(t)

t t0 0

1

-2

2

1 1

t

x(t)* ν(t)

0 1 2-1

1

-1

Solution. First, rename the time variable t into λ, say. Next, flip and shift one of thesignals. We will flip x(λ) to get x(−λ):

x(-λ) ν(λ)

λ λ0 0

1

-2

2

1 1

Now, shift by t to get x(t − λ). Note that there is no overlap between x(t − λ) and ν(λ) aslong as t ≤ −1 or t ≥ 2:

x(-1-λ) ν(λ)

λ λ0 0

1

-2

2

1 1

x(2-λ) ν(λ)

λ λ0 0

1

-2

2

1 1

6

Exam I: Solutions

So, y(t) = x(t) ⋆ ν(t) = 0 when t ≤ −1 or when t ≥ 2.

Now, we consider the remaining cases:

−1 ≤ t ≤ 0:

λ0

2

1

t t+1t-1

y(t) =

∫ t+1

0(−2) · λdλ = −

[

λ2]t+1

0= −(t + 1)2

0 ≤ t ≤ 1:

λ0

2

1t t+1t-1

y(t) =

∫ t

02 · λdλ +

∫ 1

t(−2) · λdλ =

[

λ2]t

0−

[

λ2]1

t= t2 − 1 + t2 = 2t2 − 1

1 ≤ t ≤ 2:

λ0

2

1t t+1t-1

y(t) =

∫ 1

t−12 · λdλ =

[

λ2]1

t−1= 1 − (t − 1)2 = −t2 + 2t

Overall,

y(t) = x(t) ⋆ ν(t) =

0, t ≤ −1−(t + 1)2, −1 ≤ t ≤ 02t2 − 1, 0 ≤ t ≤ 1−t2 + 2t, 1 ≤ t ≤ 20, t ≥ 2

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