# Why does adding a time delayed version of a signal to itself create a filtered signal?

I was asked this question and couldn't come up with an answer on the spot which didn't involve the frequency domain (basically that the co-efficients of the delay sequence are the impulse response of a FIR filter).

Does anyone have any insight which makes this process 'obvious'?

When you delay a signal by $T$ seconds and add it to the signal itself, you are cancelling out or nulling the signal component at frequency $\frac{1}{2T}$ Hz since that signal component will have changed phase by exactly $\pi$: \begin{align} \sin\left(2\pi\frac{1}{2T}t + \theta\right) + \sin\left(2\pi\frac{1}{2T}(t-T) + \theta\right) &= \sin\left(2\pi\frac{1}{2T}t + \theta\right) + \sin\left(2\pi\frac{1}{2T}t + \theta - \pi\right)\\ &= \sin\left(2\pi\frac{1}{2T}t + \theta\right) +\sin\left(2\pi\frac{1}{2T}t + \theta\right)\cos(\pi)\\ &\ \hspace{0.2in}-\cos\left(2\pi\frac{1}{2T}t + \theta\right)\sin(\pi)\\ &= \sin\left(2\pi\frac{1}{2T}t + \theta\right) -\sin\left(2\pi\frac{1}{2T}t + \theta\right)-0\\ &= 0. \end{align} A similar thing happens at odd multiples of $\frac{1}{2T}$ Hz also. For nearby frequencies, the cancellation is not as complete, and of course, at even multiples of $\frac{1}{2T}$ Hz, the signal component is doubled in value instead of being cancelled. Similarly, if the delayed signal is reduced in amplitude, cancellation is not complete at $\frac{1}{2T}$ Hz etc.

To summarize, the signal is being filtered because different frequencies are being passed through with different gains.

If you want the frequency-domain explanation, the transfer function $H(f)$ of the system is the Fourier transform of what Matt's answer gave as the impulse response, viz. $$\mathcal F\left[\delta(t) + \delta(t-T)\right] = 1+\exp(-j2\pi fT)$$ which is a nonconstant function of $f$ (in fact, $|H(f)|$ varies sinusoidally from a maximum of $2$ to a minimum of $0$ as discussed above), and so $Y(f)=H(f)X(f)$ is not a scalar multiple of $X(f)$. Filtering!

• Sorry for the delay - how would I go from here (that filtering is interference) to the necessity that filtering is convolution of the two signals? I can see it (algebraically) from the sum of two cosines formula, but I can't intuit a reason why. – Tom Kealy May 2 '13 at 15:07
• Please explain what you mean by "filtering is interference". I don't understand this notion at all – Dilip Sarwate May 2 '13 at 21:31
• Well, we've just established (or have we?) that adding two signals together with different phases is equivalent to filtering with a time delay because the waves interfere. How would I go (in the time domain) from there to convolution? – Tom Kealy May 3 '13 at 17:24
• I still don't understand the question. $x(t)+x(t-T) = y(t)$ is the output of a filter with impulse response $h(t) = \delta(t)+\delta(t-T)$ whose input happens to be $x(t)$, as has been pointed out in Matt's answer. If you want to write the output as a convolution, you can write $$y(t) = x\star h = \int_{-\infty}^\infty x(t-u)h(u)\,du = \int_{-\infty}^\infty x(t-u)[\delta(u)+\delta(u-T)]\,du$$ where, when you evaluate the integrals using the sifting property of impulses, you get $x(t)+x(t-T)$ which you knew already. – Dilip Sarwate May 3 '13 at 20:00

If you define (linear time-invariant) filtering as convolution, then the answer is obvious: the sum of a signal and a delayed version of it can be written as a convolution with an impulse response $h(t)$: $$h(t) = \delta(t) + \delta(t-T)$$ where $T$ is the delay between the two versions of the signal.

If the time delay of the delayed added version of a signal is exactly one cycle of any periodic content, then the output will be additively increased. If the delay is exactly half the period of any sinusoidal component, then that component will destructively interfere, and thus be zero-ed out of the output. If the delay is zero, then the signal will be doubled. For frequency/phase combinations that are between complete destructive interference or complete addition, the additive result will also be in between.

Increasing and decreasing the output depending on the frequency content of the input is typical filtering.