Daniel Townsend

3.1

To begin, it is useful to determine whether the function is even or odd, since this will eliminate half the calculations youneed to do.

. . . . . .

0

1

−2T −1T 1T 2T

It’s pretty clear that this signal is odd.

Before we start, it is helpful to get the formula for the function over 1 period, whichever is the easiest to work with. Inthis case, the period from 0 < t < T is the easiest.

x(t) =t

Tfor 0 < t < T

Since this function is odd, it means that an = 0 ∀ n = 1, 2, 3, . . .

But as you can see, a0 is not included in this. To find ao, you can use the function to find an average value over 1 periodand solve for a0

a02

=1

T

T∫0

x(t)dt

∴ ao =2

T

T∫0

x(t) dt =2

T

T∫0

t

Tdt =

2

T 2

T∫0

t dt =2

T 2

[t2

2

]T0

=2

T 2

T 2

2= 1

∴ a0 = 1

Now we need to solve for bn, which is governed by the following equation.

bn =2

T

T∫0

x(t) sin(nω0t) dt where ω0 =2π

T

∴ bn =2

T

T∫0

x(t) sin

(2πn

Tt

)dt

Let’s solve this.

bn =2

T

T∫0

t

Tsin

(2πn

Tt

)dt =

2

T 2

T∫0

t sin

(2πn

Tt

)dt

Integration by parts is necessary here.

1

Daniel Townsendu = t dv = sin

(2πn

Tt

)dt

du = dt v = − T

2πncos

(2πn

Tt

)

=2

T 2

−t T2πn

cos

(2πn

Tt

)+

T∫0

T

2πncos

(2πn

Tt

)dt

=2

T 2

[−t T

2πncos

(2πn

Tt

)+

T 2

(2πn)2sin

(2πn

Tt

)]T0

=2

T 2

[− T 2

2πncos (2πn) +

T 2

(2πn)2sin (2πn)

]

Since n is only of integer value, sin (2πn) = 0 and cos (2πn) = 1

=2

T 2

[− T 2

2πn

]

The T 2 and 2 cancel out, leaving us with

bn = − 1

πn

2

Daniel Townsend

Note that all the colors of the variables in the three convolution examples are just used as an aid in understanding variablesubstitution. The colors can be removed and it still makes sense, it’s just easier to see when you’re substituting in the samevariable with itself shifted by a constant.

6.3

f1(t)

1 2

1

0

f2(t)

1 2 3

1

0

You need to transform f2(t) so it is reversed and along the y-axis.

Do this by first adding 1 to the inside

f2(t+ 1)

1 2

1

0

Then making it negative on the inside

f2(−(t+ 1))

−2 −1

1

0

To convolve them together, start where they first meet and continue until you reach a section where the formula for eitherof the lines changes.

You need to know the equations of the lines for each segment you are working with.

Here f2(τ) is always equal to 1, while f1(τ) =

{τ 0 < τ < 11 1 < τ < 2

Let’s start with the first region, defined where 0 < τ < 1

tt-2−2 −1 1 2

1

0

3

Daniel Townsendf1(t) ∗ f2(t+ 1) =

t∫0

(τ)(1)dτ =

t∫0

τdτ =τ2

2

∣∣∣∣t0

=t2

2for 0 < t < 1

Now for the second region we will have to do two integrals, one for the part of the box in the region from 1 < τ < 2 andone for the part of the box in 0 < τ < 1

tt-2−2 −1 1 2

1

0

f1(t) ∗ f2(t+ 1) =

1∫0

(τ)(1)dτ +

t∫1

(1)(1)dτ =

1∫0

τdτ +

t∫1

1dτ =τ2

2

∣∣∣∣10

+ τ |t1

= t− 1

2for 1 < t < 2

The third region is when t is no longer in the bounds of integration, but we now have to deal with t− 2

tt-2−2 −1 1 2 3

1

0

f1(t) ∗ f2(t+ 1) =

1∫t−2

τdτ +

2∫1

1dτ =τ2

2

∣∣∣∣1t−2

+ τ |21

=3

2− (t− 2)2

2for 2 < t < 3

Now the last region is the only one left, this is where 3 < τ < 4 and even though t wont be in there at all, the tail endstill is!

tt-2−2 −1 1 2 3 4

1

0

4

Daniel Townsendf1(t) ∗ f2(t+ 1) =

2∫t−2

(1)(1)dτ = τ |2t−2

= t− 4 for 3 < t < 4

The entire piecewise function for f1(t) ∗ f2(t+ 1) can now be written.

f1(t) ∗ f2(t+ 1) =

t2

2 0 < t < 1

t− 12 1 < t < 2

32 −

(t−2)22 2 < t < 3

t− 4 3 < t < 4

0 otherwise

Using the Time Shifting property of convolution, we know that f1(t)∗f2(t+ 1) = f(t+ 1). However, we want to get f(t) !

To do this, make a variable substitution in the piecewise function where t = t+ 1 and t− 1 = t

f(t) =

(t−1)22 0 < (t− 1) < 1

(t− 1)− 12 1 < (t− 1) < 2

32 −

((t−1)−2)22 2 < (t− 1) < 3

(t− 1)− 4 3 < (t− 1) < 4

0 otherwise

Solving this gives you

f(t) =

(t−1)22 1 < t < 2

t− 32 2 < t < 3

32 −

(t−3)22 3 < t < 4

t− 5 4 < t < 5

0 otherwise

Note that in Gajic’s textbook the answer he has underneath the problem statement is wrong, since he substituted int = t+ 2 and t− 2 = t (most likely an accident)

5

Daniel Townsend

6.4

t

f1(t) =

{t 0 < t < 10 otherwise

1 2

1

0t

f2(t) =

{t− 1 1 < t < 20 otherwise

1 2 3

1

0

We have to add 1 from the inside f2(t) to get it next to he origin. To do this, substitute t = t− 1 and t+ 1 = t

t

f2(t+ 1) =

{t 0 < t < 10 otherwise

1 2

1

0

We now flip it over the axis by making the inside negative.

−t

f2(−(t+ 1)) =

{−t −1 < t < 00 otherwise

−1−2

1

0

Note that we need to get f2 in terms of t− τ where our τ = t. We therefore substitute t− (τ + 1) in for into the piecewisefunction to get the following equation. We also need to get f1(t) in terms of tau. Simply substitute in τ for t and you haveit.

f2(t− (τ + 1)) =

{t− τ (t− 1) < τ < t0 otherwise

f1(τ) =

{τ 0 < τ < 10 otherwise

Having done this, we can now convolve f1(t) and f2(t+ 1)

t−1 1 2 3

1

0 tt− 1

f1(t) ∗ f2(t+ 1) =

t∫0

f1(τ)f2(t− (τ + 1))dτ =

t∫0

(τ)(t− τ)dτ =

t∫0

τt− τ2dτ =

(τ2

2t− τ3

3

)∣∣∣∣t0

=t3

6for 0 < t < 1

6

Daniel Townsend

The next integral just so happens to be the last integral, just because there is only 1 more area where the bounds touchbefore the signals aren’t on top of each other at all.

t−1 1 2 3

1

0 tt− 1

f1(t) ∗ f2(t+ 1) =

1∫t−1

f1(τ)f2(t− (τ + 1))dτ =

1∫t−1

(τ)(t− τ)dτ =

1∫t−1

τt− τ2dτ =

(τ2

2t− τ3

3

)∣∣∣∣1t−1

t

2− 1

3− (t− 1)2

2t+

(t− 1)3

3for 1 < t < 2

We can now write out the piecewise function for f1(t) ∗ f2(t+ 1) = f(t+ 1)

f1(t) ∗ f2(t+ 1) = f(t+ 1) =

t3

6 0 < t < 1

t2 −

13 −

(t−1)22 t+ (t−1)3

3 1 < t < 2

0 otherwise

Since we have f(t+ 1) and we want f(t), we must do a final variable substitution to get the correct answer.

Here t = t− 1 and t+ 1 = t

f(t) =

(t−1)36 0 < (t− 1) < 1

(t−1)2 − 1

3 −((t−1)−1)2

2 (t− 1) + ((t−1)−1)33 1 < (t− 1) < 2

0 otherwise

Which when solved leaves you with the following.

f(t) =

(t−1)36 1 < t < 2

(t−1)2 − 1

3 −(t−2)2

2 (t− 1) + (t−2)33 2 < t < 3

0 otherwise

7

Daniel Townsend

6.5

t

f1(t)

1 2

1

0t

f2(t)

1 2

1

0

Since f2(t)’s edge is already at the y-axis, simply flip it.

−t

f2(−t)

−1

1

0

It is necessary to figure out what the equations of the functions are in what intervals.

f2(τ) = 1 That’s why we picked it to be shuffled.

f1(τ) =

{1− τ 0 < τ < 1τ − 1 1 < τ < 2

Now start shufflin’ that little bitch across f1(t)

τ−1 1 2

1

0 tt− 1

f1(t) ∗ f2(t) =

t∫0

(1− τ)(1)dτ =

t∫0

1− τdτ = τ − τ2

2

∣∣∣∣t0

t− t2

2for 0 < t < 1

The next interval will require two separate integrals, one for the head and one for the tail.

8

Daniel Townsend

τ−1 1 2

1

0 tt− 1

f1(t) ∗ f2(t) =

1∫t−1

(1− τ)(1)dτ +

t∫1

(τ − 1)(1)dτ =

(τ − τ2

2

)∣∣∣∣1t−1

+

(τ2

2− τ)∣∣∣∣t

1

= t2 − 3t+5

2for 1 < t < 2

Now there is only one integrable region left, the tail end over 1 < τ < 2

τ−1 1 2

1

0 tt− 1

f1(t) ∗ f2(t) =

2∫t−1

(τ − 1)(1)dτ =τ2

2− τ∣∣∣∣2t−1

= − t2

2+ 2t− 3

2for 2 < t < 3

We can now write the full piecewise function for f1(t) ∗ f2(t) = f(t)

f1(t) ∗ f2(t) = f(t) =

t− t2

2 0 < t < 1

t2 − 3t+ 52 1 < t < 2

− t2

2 + 2t− 32 2 < t < 3

0 otherwise

9

Daniel Townsend

7.6 d

First I got λ3 + 3λ2 + 3λ+ 1 = 0

I guessed that λ+ 1 would factor into it and did long division, shown below using the polynom package

x2 + 2x+ 1

x+ 1)

x3 + 3x2 + 3x+ 1− x3 − x2

2x2 + 3x− 2x2 − 2x

x+ 1− x− 1

0

And λ2 + 2λ+ 1 is easy to factor so we have (λ+ 1)3 = 0 to give us λ1 = λ2 = λ3 = −1

Giving us h0(t) = (C1 + tC2 + t2C3)e−t

From equation 7.41 we know that

h0(0+) = 0 = C1

h′0(0+) = 0 = C2

h′′0(0+) = 1 = 2C3 ∴ C3 =1

2

h0(t) =1

2t2e−t

h′0(t) =1

2

[−t2 + 2t

]e−t

h′′0(t) =1

2

[t2 − 4t+ 2

]e−t

h(t) =[t2 − 4t+ 2

]e−t +

1

2

[−t2 + 2t

]e−t + t2e−t

h(t) =

[t2 − 4t+ 2− 1

2t2 + t+ t2

]e−t

h(t) =

[3

2t2 − 3t+ 2

]e−t

10

Daniel Townsend

7.10 a

solve yh(t)

λ2 + 2λ+ 1 = 0

(λ+ 1)2 Double Root

yh(t) = (C1 + C2t)e−t

solve yp(t)

F (s) =1

s+ 3

H(s) =1

s2 + 2s+ 1

F (s)H(s) =1

(s+ 3)(s+ 1)2=

k1s+ 3

+k2s+ 1

+k3

(s+ 1)2=

1

4

1

s+ 3− 1

4

1

s+ 1+

1

2

1

(s+ 1)2

L−1{

1

4

1

s+ 3− 1

4

1

s+ 1+

1

2

1

(s+ 1)2

}=

1

4e−3t − 1

4e−t +

1

2te−t

Putting them together to get y(t)

Solve for the Constants

y(0−) = 1 = C1

y′(0−) = 0 = (−1)(C1 + 0) + (C2) ∴ C2 = 1

y(t) = e−t + te−t +1

4e−3t − 1

4e−t +

1

2te−t =

1

4e−3t +

3

4e−t +

3

2te−t

11

Daniel Townsend

7.14

yh(t)

λ2 + 2λ+ 1 = 0

λ1 = λ2 = −1 Double Root

yh(t) = (C1 + C2t)e−t

yh(0−) = y(0−) = 1 = C1

y′h(0−) = y′(0−) = 2 = (−1)(C1 + C2(0))e0 + (e0)(C2)

∴ C2 − C1 = 2

∴ C2 = 3

∴ yh(t) = (1 + 3t)e−t

yp(t)

h0(t) = (C1 + C2t)e−t

h0(0+) = C1 = 0

h0(0+) = C2 = 1

h0(t) = te−t

h(t) = te−t since the numerator of the transfer function is 1

yp(t) =t∫

0−(τe−τ )(u(t− τ))dt =

t∫0−τe−τdτ

This requires integration by partsu = τ dv = e−τdτdu = dτ v = −e−τ

yp(t) = −τe−τ +t∫

0−e−τdτ = −τe−τ − e−τ |t0− = 1− te−t − e−t

y(t)

y(t) = yh(t) + yp(t) = e−t + 3te−t + 1− te−t − e−t

y(t) = 2te−t + 1

12

Daniel Townsend

7.21 a

n = 2,m = 1 ∴ h[1] = b1 = 1

h[k + 2] + 4h[k + 1] + 4h[k] = δ[k + 1] + 4δ[k]

For k = 0

h[2] + 4h[1] + 4h[0] = δ[1] + 4δ[0]

h[2] + 4 = 4

h[2] = 0

Now for characteristic equation

ρ2 + 4ρ+ 4 = 0

(ρ+ 2)2 = 0

ρ1 = −2, ρ2 = −2 Double Root

h[k] = (C1 + C2k)(−2)k

Now solve for constants

h[1] = (C1 + C2(1))(−2)1 = 1

C1 = −(C2 +

1

2

)h[2] = (C1 + C2(2))(−2)2 = 0

4C1 + 8C2 = 0

C1 = −2C2

∴ C1 = −1, C2 =1

2

Therefore

h[k] = −(−2)k +1

2k(−2)k

13

Daniel Townsend

7.21 b

n = 2,m = 1

h[n−m] = h[1] = b1 = 1

∴ h[1] = 1 & h[0] = 0

h[k + 2] +1

6h[k + 1]− 1

6h[k] = δ[k + 1] + δ[k]

For k = 0

h[2] +1

6h[1]− 1

6h[0] = δ[1] + δ[0]

h[2] +1

6= 1

∴ h[2] =5

6

Now solve the characteristic equation

ρ2 +1

6ρ− 1

6

ρ1 = −1

2& ρ2 =

1

3

h[k] = C1

(−1

2

)k+ C2

(1

3

)k

h[1] = 1 = C1

(−1

2

)1

+ C2

(1

3

)1

C2 =3

2C1 + 3

h[2] =5

6= C1

(−1

2

)2

+ C2

(1

3

)2

Solving for these we get

C1 =6

5& C2 =

24

5

Therefore

h[k] =6

5

(−1

2

)k+

24

5

(1

3

)k

14

Daniel Townsend

7.22

To solve this you need to first determine what n and m are. n = 5 and m = 3.

Therefore h[n−m] = h[5− 3] = h[2] = bm = b3 = 1

∴ h[2] = 1

And we also know that h[1] = 0 and h[0] = 0 since n−m > 1 > 0

Now substitute f [k] = δ[k] and y[k] = h[k] to get the template equation.

h[k + 5]− 3h[k + 3] + 4h[k + 2]− h[k + 1] + 2h[k] = δ[k + 3]− 2δ[k + 2] + 5δ[k + 1] + δ[k]

Sift through the different values of k to get all the initial conditions

For k = −2:

h[3]− 3h[1] + 4h[0] = δ[1]− 2δ[0]

h[3] = −2

For k = −1:

h[4]− 3h[2] + 4h[1]− h[0] = δ[2]− 2δ[1] + 5δ[0]

h[4]− 3(1) = 5

h[4] = 8

For k = 0:

h[5]− 3h[3] + 4h[2]− h[1] + 2h[0] = δ[3]− 2δ[2] + 5δ[1] + δ[0]

h[5]− 3(−2) + 4(1) = 1

h[5] = 1− 6− 4 = −9

15

Daniel Townsend

7.28 b

To figure out stability, you need to find the eigen values and the poles.

λ3 + 2λ2 + λ+ 2 = 0

x2 + 1

x+ 2)

x3 + 2x2 + x+ 2− x3 − 2x2

x+ 2− x− 2

0

(λ+ 2)(λ2 + 1) = 0

∴ λ1,2 = ±j and λ3 = −2

There are no double roots. The system is marginally stable since the real part of the roots are ≤ 0, however notasymptotically stable since they are equal to zero for λ1 and λ2

This is not BIBO stable since there are two eigenvalues which are imaginary.

16

Daniel Townsend

7.32 a

ρ2 + 1.5ρ− 1 = 0

ρ =−1.5±

√1.52 − 4(1)(−1)

2(1)=− 3

2 ±√

254

2(1)= −3

4± 5

4

ρ1 =1

2, ρ2 = −2

Since the one eigenvalue is outside the unit circle, this is unstable.

Also it is not BIBO stable since the eigenvalues are not in the unit circle.

17

Daniel Townsend

12.9 a

M1(s) =25

s2 + 8s+ 25=

ω2n

s2 + 2ςωns+ ω2n

From this it’s pretty easy to see that ωn =√

25 = 5

Also we can see that 8 = 2ςωn

Plugging in for ωn we can solve for ς

ς =8

10=

4

5

ς is really the only variable we need to determine the MPOS.

MPOS = e−

(ςπ√

1− ς2)

= e

−

45π√

1− 45

2

If you plug that shit into your calculator you’ll get about 1.52%

To determine t5%s , use the equation in the book.

t5%s ≈ 3

ςωn=

345 · 5

∴ t5%s ≈ 3

4

18

Daniel Townsend

12.10

Since H(s) =1

τ1τ2

s2 +(

1τ1

+ 1τ2

)s+ 1

τ1τ2

N(s) = 1τ1τ2

and D(s) = s2 +(

1τ1

+ 1τ2

)s+ 1

τ1τ2

M(s) =N(s)

N(s) +D(s)=

1τ1τ2

s2 +(

1τ1

+ 1τ2

)s+ 2

τ1τ2

You need the numerator to be the same as the last term of the denominator, so factor out a 12

M(s) =

(1

2

)2N(s)

N(s) +D(s)=

1

2

2τ1τ2

s2 +(

1τ1

+ 1τ2

)s+ 2

τ1τ2

Use the equations to find ωn and ς

ωn =√

2τ1τ2

ς =

(1τ1

+ 1τ2

)2√

2τ1τ2

19

Daniel Townsend

12.13

G(s) = 1 (unity feedback)

H(s) =114.26s2 + 1535.49s+ 3592.09

s4 + 24.32s3 + 151.92s2

For Unit Step Error

Kp = lims→0{G(s)H(s)} = lim

s→0

{114.26s2 + 1535.49s+ 3592.09

s4 + 24.32s3 + 151.92s2

}=∞

estepss =1

1 +Kp= 0

For Unit Ramp Error

Kv = lims→0{sG(s)H(s)} = lim

s→0

{114.26s2 + 1535.49s+ 3592.09

s3 + 24.32s2 + 151.92s

}=∞

erampss =

1

Kv= 0

For Unit Parabolic Error

Ka = lims→0{s2G(s)H(s)} = lim

s→0

{114.26s2 + 1535.49s+ 3592.09

s2 + 24.32s+ 151.92

}=

3592.09

151.92≈ 23.6446

eparabolicss =2

Ka=

2

23.6446= 0.0846

20