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I don't understand the change of variable in this summation $$ \frac{1}{T}\sum_{k=-\infty}^{\infty} H \Big (\frac{F-k}{T}\Big )=1 $$ Change of variable: $f=\frac{F}{T}$ so $$ \frac{1}{T}\sum_{k=-\infty}^{\infty} H \Big (f-\frac{k}{T}\Big )=1 $$ Isn't this wrong? $T$ isn't changed?

Shouldn't it instead be $$ \frac{f}{F}\sum_{k=-\infty}^{\infty} H \Big (f-\frac{f}{F}k\Big )=1 \quad \text{?} $$

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    You express surprise that $T$ isn't changed, but in your proposed form, one can wonder why $F$ isn't changed. – 2017-01-20
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    The change of variable $f=\frac{F}{T}$ is correct. What you are doing is $T=\frac{F}{f}$. – 2017-01-20
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    @WiCK3DPOiSON : In traditional mathematical usage, the equality $a=\dfrac b c$ is logically equivalent to $c=\dfrac b a$ (at least when $a\ne0\ne c$). This is one situation where it is useful to write $f:=\dfrac F T$ or $T:=\dfrac F f.$ Those are not synonymous. – 2017-01-20
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    @MichaelHardy What I meant by '=' was substituting the variable. I came across := in Pascal which is variable assignment operator.. – 2017-01-20

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In fact, both forms are correct. However when solving certain problems, not substituting the change for all variables is convenient.

You have your summation: $$\frac{1}{T}\sum_{k=-\infty}^{\infty} H \Big (\frac{F-k}{T}\Big )=1$$ The fractions may be separated to give: $$\frac{1}{T}\sum_{k=-\infty}^{\infty} H \Big (\frac{F}{T}-\frac{k}{T}\Big )=1$$ And so now, you can apply the substitution $f=\frac{F}{T}$ for only the $\frac{F}{T}$ term: $$\boxed{\frac{1}{T}\sum_{k=-\infty}^{\infty} H \Big (f-\frac{k}{T}\Big )=1}$$

Note that: $$f=\frac{F}{T} \Rightarrow T=\frac{F}{f}$$

Therefore, you can substitute for $T$ to give your version of the summation, where you do not have the summation in terms of $T$ at all.

$$\frac{1}{(\frac{F}{f})}\sum_{k=-\infty}^{\infty} H \Big (f-\frac{k}{(\frac{F}{f})}\Big )=1$$

$$\boxed{\frac{f}{F}\sum_{k=-\infty}^{\infty} H \Big (f-\frac{f}{F} k\Big )=1}$$

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    Thanks for your answer! But suppose I set $k=T-1$ ($T=k+1$), then I guess I have to change all variables? I mean correct would be $\frac{1}{k+1}\sum^{\infty}_{T=-\infty} H \Big(\frac{F-T+1}{k+1} \Big )$ and not $\frac{1}{k+1}\sum^{\infty}_{T=-\infty} H \Big(\frac{F-k}{k+1} \Big )$ or $\frac{1}{k+1}\sum^{\infty}_{T=-\infty} H \Big(\frac{F-T+1}{T} \Big )$? – 2017-01-20
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    Well, if you are considering substituting variables which are with respect to the limits you set in your summation, you should express everything in terms of that variable, otherwise $k$ will not be summated like we would expect $T$ to. Since you've set your summation bounds in this case wrt. $T$, you should not have $k$ in your expression. However, since your previous example had the substitution $f=\frac{f}{T}$, where $f$ and $T$ are not in terms of $k$ (the summation limits), it is ok to substitute it however you want. – 2017-01-20