I am having a hard time plotting this graph. Do I express it as $u[2(k - 1.5)]$? But then still the step occurs for $k = 1.5$ which isn't possible for discrete time signals as they take integer values for $k$. So how to approach this?


2 Answers 2


$u[n]$ is the unit step function with the definition

$$ u[n] = \begin{cases}{ 1 ~~~, ~~~n \geq 0 \\ 0 ~~~,~~~ n < 0 } \end{cases} $$

Then you will have for $u[2k-3]$ as

$$ u[2k-3] = \begin{cases}{ 1 ~~~, ~~~ 2k-3 \geq 0 \\ 0 ~~~,~~~ 2k-3 < 0 } \end{cases} $$

$$ u[2k-3] = \begin{cases}{ 1 ~~~, ~~~ k \geq 2 \\ 0 ~~~,~~~ k< 2 } \end{cases} $$

I think this last line is easy to see. You can plot it wrt the variable $k$.


Dealing with indices, shifts ($-3$) and scaling ($2k$) can be confusing. Most functions used in such exercices are made of combinations of basic signal functions like discrete $\delta$ Diracs, unit steps or discrete Heaviside, etc.

Hence, in such exercices, they are often piece-wise constant or piece-wise linear. So you can determine the resulting signal from the breakpoints only, and "interpolate" the remaining points.

A first option is to take a graph paper, and sketch the function around its change points: the change point (if it exist), and the closest points on the left and right.

Here, $u[2k-3]$ will be a shifted and time-scaled version of $u$, so it is piece-wise constant. It changes when $k=3/2$ (does not exist on the discrete axis). Let us check:

  • at $k = 1$ (just below $3/2$): $u[2\times 1- 3] =u[-1]=0$,
  • at $k = 2$ (just above $3/2$): $u[2\times 2- 3] =u[0]=1$.

Then, you have it, by extending the constant on each side. Same works for piece-wise linear (even polynomials). For a non-graphical clean proof, there you have @Fat32 answer.


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