In Chapter 3 of the [Goldsmith book] (https://faee.sut.ac.ir/Downloads/AcademicStaff/1/Courses/7/Andrea%20Goldsmith-Wireless%20Communications-Cambridge%20University%20Press%20(2005).pdf) we can see that the channel impulse response for a time invariant channel is given in equation (3.10) as

$$ c(\tau) = \sum_{n=0}^N \alpha_n {\exp{(-j\phi_n)}} \delta (\tau - \tau_n) .$$

If we assume that we have no motion we obtain that $\phi_n = 2\pi f_c \tau_n$ and $$ c(\tau) = \sum_{n=0}^N \alpha_n \exp{(-j2\pi f_c \tau_n)} \delta (\tau - \tau_n) .$$

Question 1: Why is the gain complex? Or in other words, why do we have the term $\exp{(-j2\pi f_c \tau_n)}$? Shouldn't the complex phase only pop up after the Fourier transform?

If we want to compute the frequency response we can use the Fourier transform and we obtain:

$$ C(f) = \sum_{n=0}^N \alpha_n \exp{(-j2\pi f_c \tau_n)} \exp{(-j2\pi f \tau_n)}.$$

Taking uniformly spaced samples over the channel bandwidth as $f_i = f_c + iB/N_c$ for $0 ≤ i < N_c$ where $B$ is the bandwidth and $N_c$ the number of carriers we obtain:

$$ C(f_i) = \sum_{n=0}^N \alpha_n \exp{(-j4\pi f_c \tau_n)} \exp{(-j2\pi \tau_niB/N_c )}.$$

Derivation from $c(\tau)$ to $C(f_i)$: We want to compute the frequency response at $f_i = f_c + iB/N_c$. Therefore we take $c(\tau)$ and we compute the frequency response at $f_i$.

$$C(f_i) = \int_{-\infty}^{\infty}\sum_{n=0}^N \alpha_n \exp{(-j2\pi f_c \tau_n)} \delta (\tau - \tau_n)\exp{(-j2\pi f_i \tau )}d\tau.$$

By exploiting the property of $\delta$, we get $$ C(f_i) = \sum_{n=0}^N \alpha_n \exp{(-j4\pi f_c \tau_n)} \exp{(-j2\pi \tau_niB/N_c )}.$$

Question 2: This formula is different from what I expected, because of the factor 4 instead of 2 in the first exponential. I know that I'm wrong somewhere, but it seems that the channel impulse response in the beginning should not have a complex gain but should just be real.

Can you please tell me what I am missing? Thank you!


2 Answers 2


That "complex gain" represents a delay. And a delay is inevitable in causal systems. The shifted impulses are just there for a channel with discrete multipath components. So that "complex gain" is absolutely necessary as it essentially represents the amount of time taken for transmission.


The OP has set a dependence on the phase parameter $\phi_n$ with the delay parameter $\tau_n$, and from that assumption the frequency response was then derived. I don't think that restriction that $\phi_n = 2\pi f_c \tau_n$ is correct.

The impulse response is the baseband equivalent impulse response, so all effects on phase from $\tau$ are already captured in the Fourier Transform of $\delta(t-\tau)$. The additional phase $\phi_n$ occurs for any asymmetry that may occur in the channel frequency response. (Without even considering multiple paths, it is easy to conceive that any one path could have a passband asymmetry due to an absorption which would lead to a complex impulse response for the channel).

We see this from the Fourier Property for causal signals that a frequency response that is complex conjugate symmetric must be real in the time domain, and similarly a frequency response that is complex conjugate anti-symmetric must be imaginary in the time domain. Therefore a complex time domain impulse response will have a frequency response that is neither symmetric or anti-symmetric.

If and only if the resulting channel (from the superposition of all channels) resulted in a complex conjugate symmetric spectrum, then and only then would all the coefficients be real.

Further details originally provided in this answer:

Given the channel impulse response as:

$$ c(\tau) = \sum_{n=0}^N \alpha_n {\exp{(-j\phi_n)}} \delta (\tau - \tau_n) .$$

This is a time domain response, as the OP understands, and with that we see the effect of $N+1$ paths combining at the receiver with three factors: gain, phase and time delay.

$\alpha_n$ is the gain scaling for path $n$.

$\exp(−jϕn)$ is the phase rotation for path $n$.

$\delta (\tau - \tau_n)$ is the time delay for path $n$.

Time delay and Phase are two different effects that easily get conflated, as for a single tone the phase and time delay can be directly related. The channel time impulse response given is the baseband equivalent impulse response, and the phase parameter is a complex rotation (not a delay).

Regarding Question 2: If the channel impulse response was real, then this would necessitate that the resulting channel distortion as a frequency response be symmetric about the carrier (complex conjugate symmetric to be accurate). A simple example of a necessary complex channel is a passband magnitude slope across the channel. And ultimately with a channel consisting of multiple reflected paths (multipath) each path can have an independent magnitude, phase and delay.

Further in signal processing we can and often do have phase shifters that are not created from time delays. In receivers the correction for time offset (timing recovery) and phase offset (carrier recovery) are done independently. A big lesson is to not get caught in the trap of confusing time delay with a phase rotation.

  • $\begingroup$ Thank you very much for your answer! However, I'm still confused. The phase rotation is explicitely defined in formula (3.3) of the Goldsmith book and it is really equal to $2\pi f_c \tau_n$ if we do not consider the Doppler and we assume a time invariant system. I really don't understand why the phase is there. Additionally, I have found that in the book Fundamentals of Wireless Communication formula (2.19) is without any phase term... $\endgroup$
    – rick87
    Commented Apr 4 at 10:15
  • $\begingroup$ @rick87 I thought I explained that so let me know your confusion. See where I say "The phase rotation for a tone at $f_c$ from the delay $\tau_n$ is $-2\pi f_c \tau_n$" and I go into detail as to how they are related. Also this other post may help you further to see how a delay in time leads to a phase for any given frequency and what that phase would be (but I think I cover that completely in the response for your question here): dsp.stackexchange.com/a/83230/21048 $\endgroup$ Commented Apr 4 at 13:08
  • $\begingroup$ Ok, thank you! Would you confirm that the formula for $C(f_i)$ is correct? $\endgroup$
    – rick87
    Commented Apr 4 at 13:36
  • $\begingroup$ It looks like you double counted index $n=0$ by combining that phase with the first term and getting the coeff 4. In any event, to confirm the result consider using sig.freqz in python or freqz in MATLAB/Octave with the complex coefficients uniformly spaced in time (make a mock example with integer spaced $\tau_n's$ to simplify and confirm your work. This will give you the magnitude and phase of the channel. Your model is assuming that the phase is constant over frequency (other than the phase given by the delay itself) which isn't realistic: every path would have an independent phase and mag. $\endgroup$ Commented Apr 4 at 14:11
  • 1
    $\begingroup$ This last comment was helpful! So basically, what we usually compute is the baseband frequency response.. Therefore we say $f_i = iB/N_c$. Hence we have a factor of 2 and not of 4... $\endgroup$
    – rick87
    Commented Apr 6 at 9:21

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