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raccourcis clavier

reference: examples for z-transform

SequenceTransform
δ(kn)\delta(k - n)znz^{-n}
11zz1\frac{z}{z - 1}
kkz(z1)2\frac{z}{(z - 1)^2}
k2k^2z(z+1)(z1)3\frac{z(z + 1)}{(z - 1)^3}
aka^kzza\frac{z}{z - a}
kakka^kaz(za)2\frac{az}{(z - a)^2}
sinak\sin akzsinaz22zcosa+1\frac{z \sin a}{z^2 - 2z \cos a + 1}
cosak\cos akz(zcosa)z22zcosa+1\frac{z(z - \cos a)}{z^2 - 2z \cos a + 1}
aksinbka^k \sin bkazsinbz22azcosb+a2\frac{az \sin b}{z^2 - 2az \cos b + a^2}
akcosbka^k \cos bkz2azcosbz22azcosb+a2\frac{z^2 - az \cos b}{z^2 - 2az \cos b + a^2}

properties

Linearity: if x(n)=af1(n)+bf2(n)x(n) = af_{1}(n) + bf_{2}(n) then X(z)=aF1(z)+bF2(z)X(z) = aF_{1}(z) + bF_{2}(z)

Time shifting:

Z[x(t)]=X(z)x(kn)=znX(z)x(k+n)=znX(z)\begin{aligned} Z[x(t)] &= X(z) \\ x(k-n) &= z^{-n}X(z) \\ x(k+n) &= z^{n}X(z) \end{aligned}

quantization error

sampling

The idea to convert analog to digital

TT is the sampling period, and 1T\frac{1}{T} is the sampling rate in cycles per second

error=M2n+1\text{error} = \frac{M}{2^{n+1}}

where nn is number of bits used for digitalisation

resolution of A/D converter

minimum value of the output that can be represented as binary number, or M2n\frac{M}{2^n}

sampled data system

reference input rr is the sequence of sample values r(kT)r(kT)

A sampler is a switch that closes every TT seconds:

r(t)=k=0r(kT)δ(tkT)(t>0)r^{*}(t) = \sum_{k=0}^{\infty} r(kT) \delta (t-kT) \quad (t>0)

Transfer function of sampled data:

R(s)=L(r(t))=k=0r(kT)eksTR^{*}(s) = \mathcal{L} (r^{*}(t)) = \sum_{k=0}^{\infty} r(kT) e^{-ksT}

definition

Let z=esTz = e^{sT}, we have the following definition:

z-transform

Z{r(t)}=F(z)=Z(r(t))=k=0r(kT)zkZ \{r(t)\} = F(z) = Z(r^{*}(t)) = \sum_{k=0}^{\infty} r(kT)z^{-k}

zero-order hold

Transfer function of Zero-Order hold

L(u(t)u(tT))=1sesTs\mathcal{L}(u(t) - u(t-T)) = \frac{1}{s} - \frac{e^{sT}}{s}

finding the discrete transfer function

G(s)=s2+4s+3s3+6s2+8sG(s) = \frac{s^2 + 4s + 3}{s^3 + 6s^2 + 8s}

G(s)=s2+4s+3s3+6s2+8s=0.375s+0.25s+2+0.375s+4G(t)=L1(G(s))=0.375+0.25e2t+0.375e4tG(z)=Z(G(t))=0.375zz1+0.25zze2T+0.375zze4T\begin{aligned} G(s) &= \frac{s^2 + 4s + 3}{s^3 + 6s^2 + 8s} = \frac{0.375}{s} + \frac{0.25}{s+2} + \frac{0.375}{s+4} \\ G(t) &= \mathcal{L}^{-1}(G(s)) = 0.375 + 0.25 e^{-2t} + 0.375 e^{-4t} \\ G(z) &= Z(G(t)) = 0.375 \frac{z}{z-1} + 0.25 \frac{z}{z-e^{-2T}} + 0.375 \frac{z}{z-e^{-4T}} \end{aligned}

inverse z-transform

G(z)x(k)G(z) \to x(k)

power series

use: when G(z) is expressed as the ratio of two polynomials in z

G(z)=a0+a1z1+a2z2+G(z) = a_{0} + a_{1} z^{-1} + a_{2} z^{-2} + \ldots

partial fraction

For example: G(z)=z(z1)(z2)=zz1+zz2=k=0(1+2k)zkG(z) = \frac{z}{(z-1)(z-2)} = \frac{-z}{z-1} + \frac{z}{z-2} = \sum_{k=0}^{\infty} (-1 + 2^k)z^{-k}

thus, g(kT)=2k1g(kT) = 2^k-1

stability

systempole location criteria on z-plane
StableAll poles inside unit circle
UnstableAny poles outside unit circle
Marginally StableOne or more poles on unit circle, remaining poles inside unit circle
poles on s-plane
poles on s-plane

mapping from s-plane to z-plane

z=eαT(cosωT+jsinωT)z = e^{\alpha T}(\cos \omega T + j \sin \omega T)

we assume s=α+jωs = \alpha + j \omega

Location on s-planeValue of α\alphaValue of eαTe^{\alpha T}Mapping on z-plane
Imaginary axis (jωj\omega)α=0\alpha = 0eαT=1e^{\alpha T} = 1On unit circle
Right half-planeα>0\alpha > 0eαT>1e^{\alpha T} > 1Outside unit circle
Left half-planeα<0\alpha < 0eαT<1e^{\alpha T} < 1Inside unit circle

final value theorem

definition

If limkx(k)\lim_{k \to \infty}x(k) exists, then the follow exists:

limkx(k)=limz1(z1)X(z)\lim_{k \to \infty} x(k) = \lim_{z \to 1} (z-1) X(z)

root locus on z-plane

  • derive open loop function KGHˉK \bar{GH}
  • Factor numerator and denominator to get open loop zeros and poles
  • Plot roots of 1+KGHˉ=01+K \bar{GH}=0 in z-plane as k varies GH(z)ˉ=N(z)D(z)\bar{GH(z)} = \frac{N(z)}{D(z)}