Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

Theory question: freq, mag response

Status
Not open for further replies.

Andy_

Newbie level 2
Newbie level 2
Joined
Aug 11, 2011
Messages
2
Helped
0
Reputation
0
Reaction score
0
Trophy points
1,281
Activity points
1,299
I am studying the book of DSP (Proakis, Manolakis) and I have a practical question about FIR filter with linear phase. In some books frequency response of a filter is H(e^jω) and magnitude response |H(e^jω)|. In this book (chapter about FIR filters with linear phase) the frequency response is formulated as H(ω) and the graphs of frequency response are the same as the graphs of magnitude response. I am complety confused :???: Can somebody explain me from practical aspect so I would understand? What is the difference between e^jω and ω?

Thank you for your help.
 

Under the precondition of linearity you are allowed to treat a sum of different sinusoidal inputs separately.
Example: Vin1: cos(wt) and Vin2=j*sin(wt).
Thus, according to Euler: Vin=Vin1+Vin2=exp(jwt).
The ouput of a linear system also is a function of exp(jwt) and - in addition - it contains a phase shift exp(j*phi).
You always can come back to Vin1 or Vin2 by using the imaginary or real part of Vin.
The advantage of this (seemingly complicated) approach is that mathematical treatment of exp(jwt) is much more simple than using sin- oder cos-function (Fourier, Lapalace, Taylor-expansion,...).
More than that, the introduction of exp(jwt) opens the way to define the complex frequency variable s=sigma+jwt as well as the complex transfer functuion H(s). The usage of these complex variables and functions has many, many advantages for analog signal processing.

Therefore, both expressions - (H(jw)) and H(exp(jwt)) - belong to the same frequency response - it is only another way to express the same thing because the frequency response H(jw) can be derived from the transfer function H(s) simply by replacing s by jw.
 

I also have another question concerning frequency and magnitude response.
In [Proakis, Manolakis] they are talking about frequency response H(ω) and in some texts about magnitude response |H(ω)|. And the pictures showing response are the same.
Attached picture is from [Proakis, Manolakis]. Isn't that a magnitude response |Hdr(ω)|?

83_1313077046.jpg
 

The answer is easy:

The frequency response H(jw) of a system consists of 2 parts: Magnitude and phase response.

As far as the figure (as shown by you) is concerned: It is a "desired" response of the magnitude (although not explicitely mentioned?). Otherwise one should see a phase (in deg or rad) on the y axis.
 

Authors use H(e^jw) to emphasize that a frequency response (magnitude and phase) is periodic with a period 2pi. This notation is adequate for discrete-time systems.
 
  • Like
Reactions: digi

    digi

    Points: 2
    Helpful Answer Positive Rating
When we write H(e^jw) we mean a function of e^jw.

But e^jw is itself a function of the variable w. Hence we may write H(e^jw) as H(w)

But to be more clear H(e^jw) is preferred.
 

Status
Not open for further replies.

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top