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Solutions to Homework

Problem 1.

Vo(s) P (s)

=

− 10 s s100 + 1

We first want to find the time-evolution, vo(t). p(t) = 100u(t), so that P (s) = (^100) s. Thus, we get

Vo(s) =

− 10 s s100 + 1

s

=

− 10

s + 1001

We inverse Laplace transform to get

vo(t) = − 10 e−^ 100 t

(A)

Time (s)

-

-

-

-

-

0

Voltage (V)

(B)

Peak value is vo(t) = −10V 99.5% of peak value:

vo(t) = 0. 995 × vo(0) − 9 .95 = − 10 × e−^ 100 t

t = − 100 × log (0.995) = 0. 501

Problem 1.

Io(s) T (s)

=

1. 5 × 10 −^2

(s + 0.3) (s + 0.05)

μA K

(A) The input, T , is a step-function in temperature with amplitude 35◦C, t(t) = 35 × u(t). Then T (s) = (^35) s. We just need to remember to add 20◦C to the final time-series. We now have

Io(s) =

35 × 1. 5 × 10 −^2

s (s + 0.3) (s + 0.05)

μA

We need to inverse Laplace transform this expression. The easiest way is probably to perform a partial fractional expansion,

Io(s) =

35 × 1. 5 × 10 −^2

s (s + 0.3) (s + 0.05)

μA

=

A

s

+

B

s + 0. 3

+

C

s + 0. 05

A(s + 0.3)(s + 0.05) + Bs(s + 0.05) + Cs(s + 0.3) s (s + 0.3) (s + 0.05)

=

s^2 A + 0. 35 sA + 0. 015 A + s^2 B + 0. 05 sB + s^2 C + 0. 3 sC s (s + 0.3) (s + 0.05)

=

s^2 (A + B + C) + s(0. 35 A + 0. 05 B + 0. 3 C) + 0. 015 A s (s + 0.3) (s + 0.05)

We can now write

A + B + C = 0

0. 35 A + 0. 05 B + 0. 3 C = 0

0. 015 A = 35 × 1. 5 × 10 −^2

We can solve these to get A = 35, B = 7, and C = −42, so that we have

Io(s) =

s

+

s + 0. 3

−

s + 0. 05 We can inverse Laplace transform this expression to get

io(t) = 35 + 7e−^0.^3 t^ − 42 e−^0.^05 t

Because we started with io(0) = 20, the correct solution is

io(t) = 20 + 35 + 7e−^0.^3 t^ − 42 e−^0.^05 t

Time (s)

Current (

μA)

(B)

t=119 s

Problem 1. We want to plot

dB = 20 log 10 |

Vo P

(jf ) |

as a function of f. Let’s insert s = jf ,

|

Vo P

(jf ) | =|

− 10 jf 100 jf + 1

| = |

− 10 jf (− 100 jf + 1) (100jf + 1) (− 100 jf + 1)

|

=|

103 f 2 − 10 jf 104 f 2 + 1

| =

√

(10^3 f 2 )^2 + (10f )^2 104 f 2 + 1

0.0001 0.0010 0.0100 0.1000 1.

f (Hz)

-

-

-

-

-

-

-

dB

We were not explicitely asked for the phase, but we can find it as

φ = tan−^1

(

Imaginary part Real part

)

= tan−^1

(

− 10 f 103 f 2

)

0.0001 0.0010 0.0100 0.1000 1.

f (Hz)

-

-

-

-

-

0

φ^

(degrees)

Problem 1. The voltage across a Josephson Junction is

EJ =

nf KJ

(in V)

We have N of them, so we get

Vo = N

nf KJ

The input quantities are N = 550, n = 97, f = 9.0646279, KJ = 483597.9, so we get

Vo = 550 ×

97 × 9. 0646279

V = 1.00000V (to 6 significant figures)

Problem 1. (A)

VQHR = NVJJ

Since VQHR = RH IX , we re-write as

IX =

NVJJ

RH

=

N 2 nfq/h^0 h q^2 k

=

Nnf 0 q^2 k 2 q =

Nnkf 0 q

(B) Rearrange solution to question A to get

N =

2 IX

nkf 0 q

Use Ix = 150 × 10 −^6 , n = 1, k = 7, f 0 = 13. 374664 × 109 , q = 1. 60217733 × 10 −^19 to get

N = 20000.

(C)

VQH = IX

h q^2 k

= 150 × 10 −^6

6. 6260755 × 10 −^34

(1. 60217733 × 10 −^19 )^27

= 0.552687V (to 6 figures)

or

VJJ =N

nf 0 2 q/h

= 20 × 103

1 × 13. 374664 × 109

2 × 1. 60217733 × 10 −^19 / 6. 6260755 × 10 −^34

=0.552688V (to 6 figures)