Notes on Linear System Analysis II - Final Examination | ECE 312, Study notes of Electrical and Electronics Engineering

Material Type: Notes; Class: Linear System Analysis II; Subject: Electrical and Computer Engineering; University: Colorado State University; Term: Spring 2008;

Typology: Study notes

Pre 2010

Uploaded on 03/10/2009

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EE 312 Linear Systems II—Final Exam Review
Test is 2 hr. open book, open notes; Calculator (no computer) Allowed. Exam is Comprehensive
(material from entire semester).
NOTE: SHOW YOUR WORK! BOX ALL ANSWERS! EXPRESS ALL LAPLACE, Z-
AND FOURIER TRANSFORMS AND TRANSFER FUNCTIONS AS RATIONAL
POLYNOMIALS!
REVIEW (Post Mid-term) (We will fill out missing info in class)
Z Transform
-One sided
n
n
ZznfnfZzFnf
=
==→
0
][])[()(][
Directly from definition (sum geometric series)
S = 1 + a + a2 + ... + an = a
a
a
n
k
n
k
=
+
=
1
11
0
IF |a| <1, S = 1 + a + a2 + ... = a
ak
k
=
=1
1
0
Thus
a
z
z
aZn
→ (Pole Locations)
- Use Tables
- Use Properties:
Linearity
Time shift (z-1 = unit delay)
Convolution;
x[n]*y[n] X(z)Y(z) =
−∞=
k
knykx ][][
-IVT: )(lim zF
z
-FVT: )(
1
lim
1zF
z
z
z
IF (z-1)F(z) has NO poles in |z| 1
- Inverse Z Transform: PFE of
z
zF )( . Use tables
- Easy if F(z) is power series (or polynomial) in z-1
- Symbolic long division (to get a few terms)
- Integral definitions (don’t use)
pf3
pf4

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EE 312 Linear Systems II—Final Exam Review

Test is 2 hr. open book, open notes; Calculator (no computer) Allowed. Exam is Comprehensive (material from entire semester).

NOTE: SHOW YOUR WORK! BOX ALL ANSWERS! EXPRESS ALL LAPLACE, Z- AND FOURIER TRANSFORMS AND TRANSFER FUNCTIONS AS RATIONAL POLYNOMIALS!

REVIEW (Post Mid-term) (We will fill out missing info in class)

Z Transform -One sided

n n

f n Z^ F z Z f n f nz

=

0

[ ] ( ) ( [ ]) [ ]

Directly from definition (sum geometric series)

S = 1 + a + a^2 + ... + an^ = a

a a

n k n

k

=

0

IF |a| <1, S = 1 + a + a^2 + ... = a

ak k

= 1

0

Thus z a

z a n Z

←→ (Pole Locations)

  • Use Tables
  • Use Properties: Linearity Time shift (z-1^ = unit delay) Convolution;

x[n]*y[n] ↔ X(z)Y(z) = ∑

=−∞

k

x [ k ] y [ n k ]

-IVT: lim F ( z ) z →∞

-FVT: ( )

lim 1 F z z

z z

→ IF (z-1)F(z) has NO poles in |z| ≥ 1

  • Inverse Z Transform: PFE of z

F ( z )

. Use tables

  • Easy if F(z) is power series (or polynomial) in z-
  • Symbolic long division (to get a few terms)
  • Integral definitions (don’t use)

LTI System Representation

ODE: Solve for y[n] y[n] – 0.9y[n-1] = 0.1x[n] Implicit Time Domain Non-zero I.C.s

TF: Y(z) = H(z)X(z) H(z) =

  1. 9

− −^1 = z

z z Explicit Time Domain Zero I.C.s

IR: y[n] = h[n]*x[n] h[n] = 0.1(0.9)n, n ≥ 0 Explicit Time Domain 0 n < 0 Zero I.C.s

System Properties: Memory (IIR/FIR) Causality (Proper) Stability (no poles in |z| ≥ 1) Invertibility (bi-proper)

Bilateral Z Transform

Fb(z) = Zb(f[n]) = n n

f n z

=−∞

∑ [ ]

f[n] ↔ Fb(z) & ROC

left-sided

right-sided

two-sided

inverse formula

  • Use tables (both LS and RS) and split signal into LS and RS pieces

Sampling Process

x(t) x[n] xp(t)

x(t) xs(t) xp(t) Impulse Modulator

Discrete Fourier Transform

X[k] = DF(X[n]) = N

j N

kn N

N

n

N

n

N j kn xne xnW W e

π 1 2 π

0

1

0

2 [ ] [ ]

− −

=

=

∑ =^ ∑^ =

x[n] = DF-1(X[k]) = N

k xnW N

kn N

N

n

[ ]

0

=

x[n] ↔ X[k]

  • numerical calculation
  • DFT ~ IDFT x[n] = NT

k T

x nW N

kn N

N

n

( [ ] )

0

=

N

kf NT

k f = = s

  • Fast Fourier Transform (FFT, IFFT) N^2 → N* ½log 2 (N) complex multiplications for complete transform