# Don't we need both negative and positive discrete complex exponentials to make a real discrete time signal?

For a continuous time periodic signal , the Fourier spectrum has both negative and positive complex exponentials in equal numbers ,but I have seen for some discrete time periodic signals it is not the case ,i.e. the discrete time Fourier series of the discrete time periodic signal does not contain discrete negative and positive complex exponentials in equal numbers. Why is it so ?

• Please supply an example of this case you site. The only one I know of is for a signal that is periodic over an even number of samples, $N$, in which case it can have a real-valued spectral "bin" at $N/2$. Commented Jul 13 at 2:24
• Yes , Even I am asking for the same case you have mentioned .Why is it so ? Commented Jul 13 at 4:52

What the OP states is true for REAL signals specifically: For any real signal in time, the spectrum (consisting of coefficients of $$e^{j\omega t}$$) will be complex conjugate symmetric (positive and negative frequency components with the same magnitude and opposite phase for any given frequency). When the spectrum is not complex conjugate symmetric, then the time domain waveform MUST be complex: every value in time will be a complex value, in that it will have a magnitude and phase and therefore can have both real and imaginary components.

This is a valuable property for use in digital wireless communications, where we are often motivated to get as much data rate between a transmitter and receiver in the minimum amount of bandwidth. If we up-convert a real baseband signal to a carrier frequency, those symmetric positive and negative frequencies as baseband will map to the upper and lower sidebands of the carrier. For the case of a complex baseband signal where the positive and negative frequencies can be independent, we can get twice the data rate in the same amount of bandwidth with all else equal over the case of a real baseband signal.

Regarding the additional question in the comments on what happens for the case of discrete signals with an even number of samples, the following two slides I have unify the interpretation of the Fourier Transform both in continuous and discrete time, and with that hopefully provide further insight on this:

In this first graphic, I introduce the Continuous Time Fourier Transform (CTFT), here for a real time domain signal, resulting in a complex conjugate spectrum (only the magnitude of this spectrum was plotted, but we note that each value in frequency also has a phase).

The CTFT is given by:

$$X(\omega) = \int_{t=-\infty}^{\infty}x(t)e^{-j\omega t}dt$$

I note here that the waveform in time is aperiodic, continuous, and the time domain (as given by the integration in the transform) extends to $$\pm \infty$$. This results in a transform that is also aperiodic, continuous, with the frequency domain $$\omega$$ extending uniquely to $$\pm \infty$$.

If we are to sample the time domain waveform, making it discrete, as introduced in the middle of the graphic above, then the frequency domain result is periodic in frequency, using the Fourier Transform of a Discrete-Time sequence, which is the Discrete-Time Fourier Transform or DTFT given as follows (where I have normalized the time index to be in samples as $$n$$ instead of seconds as $$t$$, and thus the frequency axis is normalized to be in radian frequency of radians/sample):

$$X(\omega) = \sum _{n=-\infty}^{\infty}x[n] e^{-j\omega n}$$

Note how the time domain index $$n$$ in the summation still extends to $$\pm \infty$$, so for that reason the frequency domain still remains continuous with non-zero values possible at any frequency.

Finally in the last graphic we introduce the Discrete Fourier Transform (DFT) with some interesting results. The famous "FFT" is an algorithm that will efficiently and accurately provide the DFT result, so we may see the terms DFT and FFT interchangeably to represent the same result (one is defined by the equation below and the other is an efficient algorithm to compute the same):

$$X[k] = \sum _{n=0} ^{N-1} x[n] e^{-jk \omega _o n}$$

What we are doing here is instead of using an infinite number of unique samples in the time domain as done with the prior variants of the Fourier Transform, we use a fixed number of total samples $$N$$. This is quite similar to the Fourier Series Expansion for those familiar with that, where a waveform with a fixed time duration from $$0$$ to $$T$$ in time is expanded into a series of frequency components each with a frequency that is an integer multiple of the "fundamental frequency" $$1/T$$. With the DFT we have $$N$$ samples in time at a sampling rate $$f_s$$ so the total time duration is $$T = N f_s$$ seconds, or as we did previously in normalized time units of samples the total time duration is simply $$N$$ samples. Thus the fundamental frequency given in radians per sample would be $$\omega_o = 2\pi/N$$, resulting in the final formula for the DFT as we would typically see it. With that we see the result in frequency is consistent mathematically with periodically extending (repeating) the waveform in time since the individual frequencies themselves are periodic. The same thing occurs with the DFT and this interpretation provides and intuition to help unify our understanding between the continuous and discrete time cases. The transforms whether they be of continuous or discrete time signals of a fixed length sequence over a fixed duration in time produce the similar result as the transform of that same sequence periodically extended to $$\pm \infty$$ in time: the frequency domain will be non-zero only at multiples of $$1/T$$ where $$T$$ is the total time duration of the sequence that is repeated (so in the DFT this would be the frequency given by $$f_s/N$$ where $$f_s$$ is the sampling rate and $$N$$ is the total number of samples in time). From this as well as with the DTFT we see the important Fourier property that periodicity in one domain indicates the signal is discrete in the other domain. The DFT has both things happening: it is periodic and discrete in time, and therefore it is discrete and periodic in frequency! (to be thorough with this interpretation, as a nit that can be ignored, note that "discrete" is really indicating "non-zero" because if we did extend the time domain to $$\pm \infty$$ for a periodic waveform --which means the DTFT of a periodic waveform-- it must be continuous in frequency - but it would be zero valued everywhere except the DC, fundamental and harmonics of the fundamental any of which may also be zero depending on the waveform).

So with that background in mind, we see the result for the DFT of $$N$$ samples of a real time domain waveform, and in this case specifically where $$N$$ is odd as 9 samples. The DFT result is typically presented with the frequency index $$k$$ going from $$k=0$$ to $$k=N-1$$. $$k=N$$ corresponds to be exactly the sampling rate and $$k$$ is in increments of the fundamental frequency given as $$f_s/N$$. Due to the frequency domain periodicity, we see in the graphic how the upper half of the DFT result is equally the result for the negative frequencies and overall no matter how far we extend the frequency axis, if we extend it symmetrically, we will see that the result in frequency is symmetric (complex conjugate symmetric) if the time domain waveform was real.

When the DFT is even, we have the complexity that the bin at $$k=N/2$$ lands exactly on the Nyquist bin (corresponding to half the sampling rate or $$f_s/2$$)-- but given the explanation and exercise described above that would be inconsequential--- if we extend that frequency domain result symmetrically periodically in frequency, the result will still be symmetric! There are applications where it may be useful to apply half the Nyquist bin to the positive frequencies and the other half as a complex conjugate to the negative frequencies. For an example of that, please refer to my answer for DSP.SE #82273.

• Thank you sir for answering . I was asking for discrete signals . Does this hold true for them too ? Commented Jul 13 at 3:12
• Yes it is universally applicable for all Fourier Transforms (continuous or discrete such as the FFT): real in one domain must be complex conjugate symmetric in the other domain Commented Jul 13 at 13:08
• then why does it fail for even number of samples as mentioned by Tim Wescott above ? Commented Jul 13 at 13:33
• That's not a "Fail" but an indication of how that bin should be interpreted--- with an even number of samples the Nyquist bin is equally a "postiive" frequency as well as a "negative" frequency. It lands both exactly on $+f_s/2$ and $-f_s/2$. Note when we get the FFT result, everything is presented as a "positive freuquency" extending from DC to one sample less than the sampling rate. We use 'fftshift' to move the upper half to equally be the negative frequencies, since the frequency domain for sampled signals is periodic: everything from $+f_s/2$ to $f_s$ is equally $-f_s/2$ to $0$! Commented Jul 13 at 14:00
• thanks for answering sir , can you please give me any references where I can read and study about this more . I searched in some textbooks but they haven't mentioned like this. can you suggest any textbooks or articles please ? Commented Jul 13 at 15:57