hmk_3_soln
TRANSCRIPT
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ECE 350 – Linear Systems I
Solutions to Homework #3
1. Transforming the circuit and applying voltage division:
V (s) = s1+ s
E (s) = ss +1
!"#
$%& 1+ e
' s
s' e
'2 s
s!"#
$%& = s
s +1( ) + se
' s
s s +1( ) ' se
'2 s
s s +1( )
( v(t ) = ) t ( ) ' e't u(t )+ e' t '1( )u(t ' 1) ' e' t '2( )u(t ' 2)
2.
Transforming the circuit and writing a nodal equation for V:
V ! 16 +1
s
"#$
%&'
8s+
V
1+
V ! 1
s
1
2s
= 0 ( V (s) =s2+ s +
1
16
s s +1
4
"#$
%&'
2 =
A
s+
B
s +1
4
+
C
s +1
4
"#$
%&'
2
( A = 1; B = 0;C = 12( v(t ) = u(t )+ 1
2te
! 1
4 t u(t )
3.
Transforming the circuit and writing mesh equations:
1
s I + s I +
1
2v
x
!"#
$%& + 2 I '
1
s
!"#
$%& = 0;v
x = 2 I '
1
s
!"#
$%& ( I (s) =
s
2+1
s2+ s +
1
2
I (s) =
1
2s +
1
2
!"#
$%&
s +1
2
!
"#
$
%&
2
+
1
2
!
"#
$
%&
2 +
3
2
1
2
!"# $%&
s +1
2
!
"#
$
%&
2
+
1
2
!
"#
$
%&
2 ( i(t ) = e
' 1
2t 1
2cos
t
2+
3
2sin
t
2
!"#
$%&
u(t )
4.
Transforming the circuit, since we have a single loop with one voltage sources and seriesimpedances:
I (s) =
E (s)+2
s
1+8
5s+
s
10
; E (s) =!0.5s + 2.5
s2+1
" I (s) =15s
2+ 25s + 20
s2+1( ) s + 2( ) s + 8( )
" i(t ) = e!2t ! 2e!8t + cos t + sint #$ %&u(t )
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5.
Transforming the circuit and writing a nodal equation at the top of the current source:
V (s) ! 2
s
10+
1
s + 3+
V (s)
10
s
= 0 " V (s) =!8s + 6
s s +1( ) s + 3( )
" v(t ) = 2 ! 7e!t + 5e!3t #$ %&u(t )
6. Transforming the circuit and writing KCL at the inverting input of the op amp:
0 ! s + 2
s2+ 4s +13
2+
0 ! V 4
s
= 0 " V (s) =!2 s + 2( )
s s2+ 4s +13( )
" v(t ) = ! 4
13+ e
!2t 4
13cos3t !
6
13sin 3t
#$%
&'(
)
*+
,
-.u(t )
7.
a.>> tplot=0:.01:10;
>> zplot=2*exp(-tplot)+tplot.*exp(-tplot)+.5*(tplot.^2).*exp(-tplot);>> plot(tplot, zplot, 'r')
b.>> t=[0 10];
>> zinit=[-1; 2];>> [t,z]=ode45(@prob4, t, zinit);>> hold on
>> plot(t,z(:,2))
m-file:function dz=prob4(t, z)dz=[-2*z(1)-z(2)+exp(-t); z(1)];
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they are the same
8.
a.
>> tplot=0:.01:10;
>> yplot=cos(2*tplot)+4*sin(2*tplot)-exp(-tplot).*cos(2*tplot)-4.5*exp(-tplot).*sin(2*tplot);
>> plot(tplot, yplot)
b.
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they are the same