# Adding noise to a synthetic value

Please, could someone let me know how I can add noise to a synthetic value. I have a temperature value and I need to add noise to this value and made a graph of this noise.

To be more clear with my question I have a single value of temperature of 1088.24(Kelvin), that is a result of thermodynamic calculation of temperature at the inlet of the Turbine, because it is a thermodynamic value it don't has error. I need to add noise to this value in order to see how will be the behaviour of a measurement value from a sensor (with measure error in this case noise) and plot the noise of this value. All this is new for me, I just start to learn about this but it is a little bit confuse. I really appreciate your collaboration.

Kind regards

• What properties do you want the noise to have? Commented Jan 31, 2018 at 16:19
• Hi Mark, thanks for your answer. Sorry this is new for me and I don't know about noise properties and how this will affect it. My professor just ask me to add some noise to a value of 1088.24 Kelvin and plot in excel the value of noise in small, medium and large variations.
– Aby
Commented Feb 1, 2018 at 11:02
• That is useful information for you. It indicates you should study the variability in the temperature measurement system: this will tell you the statistical nature of the relevant "noise." Without that, we cannot suggest any solution that is scientifically relevant: all answers will be arbitrary. In other words, you're basically asking us to tell you the characteristics of your sensor without giving us any information about it.
– whuber
Commented Feb 1, 2018 at 15:38

Assuming the temperature value you're talking about – let's call it $\bar\theta$ – is the mean of a sample of $n$ measurements $\theta_i$ .

So you have a model that looks like this:

$\bar\theta = \frac1n\sum\limits_{i=1}^n\theta_i$.

Then after adding some noise – let's call it $\epsilon$ – to your data you want the mean to stay unchanged. So $\epsilon$ has to cancel out and $\bar\theta$ should be the same with or without $\epsilon$ in your model. Hence, $\epsilon=0$ and your model now looks like this:

$\bar\theta = \frac1n\sum\limits_{i=1}^n\theta_i+\epsilon$.

Because $\epsilon$ is actually the mean of measurement errors $\epsilon=\bar\epsilon = \frac1n\sum\nolimits_{i=1}^n\epsilon_i$ your model now looks like this:

$\bar\theta = \frac1n\sum\limits_{i=1}^n(\theta_i + \epsilon_i)$.

As you can see, all the $\bar\theta$ are the same because of the $\epsilon_i$ ar canceling out to $0$.

Now, as @MarkL.Stone stated in the comment it depends on what properties you want $\epsilon$ to have. Often $\epsilon$ is normal distributed with the mean of $0$ and some standard deviation $\sigma$, whereby the case $\sigma=1$ is called standard normal distribution $\mathrm{N}(0, 1)$.

To visualize $\epsilon$ should now be straightforward.