Đề tài Complex Frequency

Hanoi University of Science and Technology
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Preparation for
Engineering Circuit Analysis Presentation
Chapter 12
Complex Frequency
	Teacher: Nguyen Van Binh
	Students: Vu Duc Tuan 20092951
	 Nguyen Manh Tuan 20092999	
	Class: BME-K54
Content
Introduction
Complex frequency
The damped sinusoidal forcing function
Z(s) and Y(s)
Frequency response as a function of σ
The complex frequency plane	
Natural respond and the s plane
A technique for synthesizing the voltage ratio H(s)=Vout/Vin
INTRODUCTION
When faced with time-varying sources, or a circuit with switches
installed, we have several choices with respect to the analysis approach. Unfortunately, not all sources are sinusoidal, and there are times when both transient and steady-state responses are required. In such instances, the Laplace transform proves to be a highly valuable tool. Many textbooks simply launch straight into the Laplace transform integral, but this approach conveys no intuitive understanding. For this reason, we have chosen to first introduce what may strike the reader at first as a somewhat odd concept—the notion of a “complex” frequency. Simply a mathematical convenience, complex frequency allows us to manipulate both periodic and nonperiodic time-varying quantities in parallel, greatly simplifying the analysis.
 COMPLEX FREQUENCY
Consider given by 
where K is a complex constant.
Here, is called the complex frequency
a) DC Case ( ) 
b) Exponential Case ( )
c) Sinusoidal Case ( )
Note that:
d) Exponentially Damped Sinusoidal Case
3) THE DAMPED SINUSOIDAL FORCING FUNCTION
The general exponentially varying sinusoid, which we may represent with the voltage function
can be expressed in terms of the complex frequency s by making use of Euler’s identity as before :
Substitute	 => 	 (forcing function)
the response may be assumed to be
	 or
the resultant forced response is complex, 
and it must have as its real part the desired time-domain forced response:
Example: Apply the forcing function to the series RLC circuit, and specify the forced response by finding values for 
 in the time-domain expression 
	Ans :
4) Z(s) and Y(s)
It is necessary to know the constant of proportionality between the complex voltage across an element and the complex current through it. This proportionality constant is the impedance or admittance of the element; it is easily determined for the resistor, inductor, and capacitor. 
Let us consider the inductor carefully and then merely present the results 
for the other elements without proof. Suppose that a voltage source 
 is applied to an inductor L; the current response must have the form 
If we present the voltage as: 
and the current as: 	
then the substitution of these expressions in the defining equation of an 
inductor, 
v(t) = L (di(t) /dt) 
leads to V=sLI è Z(s) = V/I = sL
In a similar manner, the admittance of an inductor L is: Y(s) = 1/Sl
The impedances of a resistor and a capacitor at a complex frequency s are obtained by a similar sequence of steps. Without going through the details, the results for all three of the elements are shown in the following table:
R
L
C
Z(s)
R
sL
1/sC
Y(s)
1/R
1/sL
sC
5) FREQUENCY RESPONSE AS A FUNCTION OF σ
Before we discuss the more general problem of frequency response as a function of the complex frequency s in the next section, let us devote a little time to the simpler problem of frequency response as a function of a. This is a subject which is very useful in synthesizing RC and RL circuits, a popular topic in more advanced courses. As a simple example we may select the series RL circuit excited by the frequency-domain voltage source Vm 0 . The current is obtained as a function of s by ividing the source voltage by the input impedance:
6) THE COMPLEX-FREQUENCY PLANE
Complex frequency has two components (σ and ω):
+ Since ω represents an oscillating function, there is no 
physical distinction between a positive and negative frequency.
+ In the case of σ, however, which can be identified with an 
exponential term, positive values are increasing in magnitude, 
whereas negative values are decaying.
Pole-Zero Constellations
+ These comments may be illustrated by considering the configuration of the poles and zeros, sometimes called a pole-zero constellation, that locatesall the critical frequencies of a frequency-domain quantity
+ A pole-zero constellation for an example impedance isshown in Fig. 15.32; in such a diagram, poles are denoted by crosses andzeros by circles.
Let us now build up the expression for Z(s) that leads to this pole-zero configuration. The zero requires a factor of (s + 2) in the numerator, andthe two poles require the factors (s + 1 − j5) and 
(s + 1 + j5) in thedenominator. Except for a multiplying constant k, we now know the form
ofZ(s):
In order to determine k, we require a value for Z(s) at some s other than a critical frequency. For this function, let us suppose we are told Z(0) =1.
By direct substitution ,we find that k is 13, and therefore
7) NATURAL RESPONSE AND THE s PLANE
Let us introduce the method by considering 
the simplest example, a seriesRL circuit as shown in Fig. 15.33. A general voltage source vs(t) causes the current i (t) to flow after closure of the switch at t = 0. The completeresponse i (t) for t >0 is composed of a natural response and a forcedresponse:
We may find the forced response by working in the frequency domain, assuming, of course, that vs(t) has a functional form that we can transform tothe frequency domain
Next we consider the natural response. From previous experience, we know that the form will be a decaying exponential with the time constant L/R, but let’s pretend that we are finding it for the first time. The form ofthe natural (source-free) response is, by definition, independent of the forcing
function; the forcing function contributes only to the magnitude of the natural response.To find the proper form, we turn off all independent sources; here, vs(t) is replaced by a short circuit. Next, we try to obtain thenatural response as a limiting case of the forced response.
Let us inspect this new idea from a slightly different vantage point.We designate this ratio H(s) and define it to be the circuit transferfunction. Then,
We see that the pole of the transfer function 
occurs at s = −R/L + j0, as shown in Fig. 15.34.
So, we thus obtain the natural response
Where A is an unknown constant. We next desire to 
transform this naturalresponse to the time domain.
andA may be determined once the initial conditions are specified for this circuit. The forced response if(t) is obtained by finding the inverse Laplacetransform of If(s).
A More General Perspective
Figure 15.35 shows single sources connected to networks containing no independent sources. The desired response, which might be some current I1(s) or some voltage V2(s), may be expressedby a transfer function thatdisplays all the critical frequencies. To be specific, we select the response V2(s) 
The poles of H(s) occur at s = s1, s4, . .and so a finite voltage V2(s) ateach of these frequencies must be a possible functional form for the naturalresponse. The natural response that occurs when theinput terminals are short-circuited must therefore have the form
8) A TECHNIQUE FOR SYNTHESIZING THE VOLTAGE RATIO H(s) = Vout/Vin
Much of the discussion in this chapter has been related to the poles and zeros of a transfer function. We have located them on the complexfrequencyplane, we have used them to express transfer functions as ratiosof factors or polynomials in s, we have calculated forced responses fromthem, and in Sec. 15.7 we have used their poles to establish the form of thenatural response.
Now let us see how we might determine a network that can provide a desired transfer function. We consider only a small part of the general problem,working with a transfer function of the form H(s) = Vout(s)/Vin(s), asindicated in Fig. 15.41. For simplicity, we restrict H(s) to critical frequencies
on the negative σ axis (including the origin). Thus, we will consider transfer functions such as
Let us begin by finding the voltage gain of the network of Fig. 15.42,which contains an ideal op amp.We therefore may set the sum of the currents entering theinverting input terminal equal to zero:
In Fig. 15.43a, we let Z1= R1, while Zfis the parallel combination of Rf and Cf. Therefore,
We have a transfer function with a single (finite) critical frequency, a pole ats = −1/Rf Cf.
Moving on to Fig. 15.43b, we now let Zf be resistive while Z1 is an RC parallel combination:
The only finite critical frequency is a zero at: 
s = −1/R1C1