Unwind di Power Electronics - 3 di 5

Example: free response calculation for a first-order system

• One real characteristic root
• One exponential mode
• General expression of the free response:
• Determination of A:
• Expression of the free response:

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For example, let's consider the following general solution:

04 iYo k K i

The associated characteristic equation is:

044 O 0K K zi

In this case there is only one real root, so one exponent mode:

Aa K 0ap

The general expression of the free response becomes:

kata4 0K

Now it must be determined the constant A so it must be imposed the initial condition:

041 I400 041a

The specific expression of the free response is:

44 0i KraK

Free response, case N<M

When N<M the free response y [k] is a superposition of system modes, plus an initial sequence of length M−N:

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Until now it has been considered the case in which N>=M, but in general it can happen the opposite case. In the case that N<M. In this case the solution of the system is a superposition of the system modes plus an initial sequence of length M-N:

k.iti i

This is a superposition of Kronecker pulses with amplitude qi. This summation is related to the case in which the input affects the output for a longer time then the output itself. This situation doesn’t appear in continuous time.

In a special case, N=0 and this case is the case of the FIR (Finite Impulse Response) systems. A FIR system is exclusively related to the discrete-time systems and is a typology of systems in which the free response goes to 0 in a finite time. The free response of a continuous-time system goes to 0 mathematically with an infinite time.

Example: free response of a first-order system with N<M

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For example, let's consider the following response of a system:

0,4 i4 itb.tltbonktbmk K 2k K i

In this case is clear that: N=1, M=2, so this is the case in which: N<M:

iµilke 22 LI i4 4µilke iI ii i 4 4 iO 0,40 tbill.ito il 0 belli0ii 1

Note that it must be evaluated the free response, so after k=0 the input is equal to 0.

iQIY.itabill.itaiball.ati 4 4 belliO Ii µii Ii i 4 4 af4 itaibillOilii 2 222itclibrll.ztaiball.tk

These last two expression can be written as follow:

i iditi4 4 tbill.itO I bzll.atI il I1 ai Mii iai4ai4 4 itbill.itbzll.at0il 22ii 22 mi

The terms into parenthesis are equal, so the general y[k]=y0[k] is:

atti Oi4 tbiltitbzll.at KK mi

Note that it’s important to impose k>0 because for k=0 the previous expression is not true anymore due to this term that doesn’t appear in the expression for k=0.

This term can be added to the previous expression in just multiplying it by a Kronecker pulse:

s iai44 itbill.itbzll.atK kmi mi

Now this expression is valid for every value of k.

Lesson #16

Lesson #17

The transfer function

• Since the forced response y [k] is the discrete-time convolution between the input u[k] and the impulse response h[k], its Z-Transform y (z) is fH(z), i.e. the Z-Transform of the impulse response, is the transfer function of the system.

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A transfer function in the context of discrete-time system is the same is the same of as in the continuous-time system. Since the forced response yf[k] is the discrete-time convolution between the input u[k] and the impulse response h[k], its Z-Transform yf(z)is:

Ye Hz µz za

Where h[k] is the system response to a Kronecker delta. Its Z-transform is H(z) and it’s called transfer function.

Transfer function expressions

• Given the time-domain difference equation the system transfer function can be expressed in terms of the a 's and b 's asi i
• Equivalently:

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To derive the expression of the transfer function H[z] it’s used the time-domain difference equation:

Superposition of delayed signals

Firstly it’s applied the Z-transform to right and left sides, successively the linearity and at least the delay theorem, whose fundamental result is the following:

HK 0ko èkoK z

Z-Transform

The general delayed signal is also causal and the time-domain difference equation is the superposition of 2 causal signals that are y[k] and u[k]. These 2 last sequences are causal because it’s calculated the forced response which means that the initial conditions are 0, for both yf and u.

Applying the Z-transform, the 2 signals becomes:

Ye4 È ÈHK Kµz zi i

After this last transformation, solving for H(z), the result is:

The transfer function is a ratio between 2 polynomials in the variable z^-1. In an equivalent way:

This is equal in respect of the previous form but it is in the variable z. Note that if: N<M, then this term is a pole and not a zero. Using the initial value theorem, it can be determined h[0]:

h Hzboiling0

Transfer function expressions

• When H(z) is written in this form, its denominator is the system's characteristic polynomial
• Assuming no pole/zero cancellation occurs, nonzero poles of H(z) coincide with the system's characteristic roots

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If denominator is imposed equal to 0, it is obtained the characteristic equation of the system, so the non-zero poles of H(z) are the system’s characteristic roots. Moreover if: N<M, then there is an additional pole at the origin whose multiplicity depends by the difference between N and M.

Stability

• System is asymptotically stable if and only if all poles are inside the unit disk
• System is marginally stable if there are no poles outside the unit disk, and there exist simple poles belonging to the unit circle (i.e. poles of magnitude one)
• System is unstable in all other cases

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The system is asymptotically stable if and only if all the poles are inside the unit disk:

Note that pi is a real pole and ri is a complex conjugate pole.

In all others cases, the system is unstable. There is at least one mode that does not remain bounded.

Transfer function expressions

• Partial fraction expansion of previous expressions:

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The transfer function can be rewritten evidencing the real and imaginary poles:

MN

The first summation considers the real roots of the characteristic equation while the second summation consider the complex and conjugate roots. This expression can be confronted with the following:

This is the equivalent time-domain expression of the transfer function H(z).

In the case that N is equal or less than M, the transfer function becomes:

MN

Note that:

M MN Nti riskf I ii ioo

Example: transfer function of a discrete-time differentiator

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Let’s evaluate the Z-transform of a differentiator starting from the time-domain definition of y[k]:

Y iKdk M MK K ihkekdfk.sk 1H E1KdZ

It can also be determined the transfer function of the integrator:

Y 4 Kimk K KI ihish hk k kIH11 Kiz Z zH KiZ E1

Example: response of a first-order system to a sinusoidal signal

3g

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Let’s consider the case of the response of a first-order system to a sinusoidal signal. The system and the input are:

Note that the normalised angle theta is defined using the normalised angular frequency wm:

a WantsÈ eiUmtsA

Lets calculate the forced response:

io bene ipae'iobaie4e k Su

The term Sk is the generic term of a geometric series. In fact if that summation goes from 0 to infinity the summation becomes a geometric series:

0httSu _a e ioal aJkHt.ci beik0 akttel0i i0igYe faeK aal1Sujk ja httbeb c aaio _alide

The second one is the transient that goes to 0.

The first term of this last result is an oscillatory steady-state, in fact it depends by a complex exponential with the same frequency but different phase and different amplitude due to this quantity, that is the frequency response. This term can also be seen as follow: