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.

Impulse response at ideal channel

Status
Not open for further replies.

grom88

Newbie level 1
Joined
Feb 18, 2012
Messages
1
Helped
0
Reputation
0
Reaction score
0
Trophy points
1,281
Activity points
1,288
Hi all! I need the value of h(t) (impulse response) in an ideal comunications channel
Its h(t) = delta(t)?
I would really appreciate a response


Thanks a lot!
 
Last edited:

what is the transfer function of the com channel?
 

Depends on your definition of an ideal channel. I guess a time delay won't endanger ideality...
 

What means ideal channel? Well, the output of an ideal channel may have some time delay compared with the input, and it may have a different amplitude than the input (just a scale change), but otherwise it must have the same shape as the input (i.e., no distortion). In other words, an ideal channel is a channel which allows distortionless transmission.

Mathematically, if x(t) is the input of an ideal channel, then its output y(t) is expressed as

y(t) = Kx(t-t0) (1)

where the constant K represents the amplitude scale and the constant t0 represents the time delay.

Taking the Fourier transform of both sides in Eq. (1) we obtain

Y(f) = KX(f)e-j2πft0 (2)

But we know that the output y(t) of a channel is the convolution of the input x(t) with its impulse response x(t),

y(t) = x(t) * h(t) = h(t) * x(t) (3) (commutative property)

and from the convolution theorem of the Fourier transform we have

Y(f) = H(f)X(f) (4) (convolution in the time-domain becomes multiplication of the corresponding spectrums in the frequency domain)

Comparing Eq. (2) and Eq. (4) we obtain the frequency response (also called transfer function) of the channel.

H(f) = Ke-j2πft0 (5)

We note that an ideal channel has a constant magnitude response (in order to not introduce magnitude distortion) and its phase shift is a linear function of frequency (in order to not introduce phase distortion).

We know that

delta(t) <-> 1 (6), or equivalent Kdelta(t) <-> K (7)

(where <-> denotes Fourier transform pair: at the left side is the time-domain signal and at the right side is its Fourier transform, or, if you want, at the right side is the frequency-domain signal and at the left side is its inverse Fourier transform)

Thus, for no magnitude distortion, a channel must have an impulse response h(t) = Kdelta(t), which implies a constant frequency response H(f) = K. But remember the following (something that many students forget): it is not enough for a channel to amplify or attenuate all frequency components of the input signal the same to be classified as an ideal channel, but these frequency components must also arrive at the output with identical time delay in order to add up correctly.

The above mentioned channel is ideal because it implies an infinite bandwidth and an impulse response with infinite time-duration. In practice, the impulse response will be time-limited and the frequency response will be band-limited.
 
  • Like
Reactions: FvM

    FvM

    Points: 2
    Helpful Answer Positive Rating
Status
Not open for further replies.

Similar threads

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top