# How to plot the discrete time signal $u[2k - 3]$?

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?

$$u[n]$$ 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 $$3/2$$): $$u[2\times 2- 3] =u=1$$.

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.