Hi everybody, I have a little trouble issues with my function handles.
In particular I am trying to describe a function handle which has a specific value (related to two variables) in a range of one of the two vairables, and 0 otherwise.
E_2rad=0.79;
@(freq_wave,wave_angle) (0<= wave_angle <= pi-E_2rad) .* ((2.25/4)*ro_w*g*B*0.75).*(sin(E_2rad+wave_angle).^2+(2.*freq_wave.*(vel_nave/g).*(cos(E_2rad).*cos(E_2rad+wave_angle)...
-cos(wave_angle)))) + (wave_angle > pi-E_2rad) .* 0
the function written as above mentioned returns a value different from 0 if wave_angle > pi-E_2rad instead and I cannot understand why and how to fix it. Thanks in advance for your help

 Akzeptierte Antwort

Dyuman Joshi
Dyuman Joshi am 21 Nov. 2023

0 Stimmen

You need to define multiple conditions separately/individually i.e -
This
(0<= wave_angle <= pi-E_2rad)
should be -
(wave_angle >=0) && (wave_angle <= pi-E_2rad)

4 Kommentare

Gianluca Angelini
Gianluca Angelini am 21 Nov. 2023
Hi Dyuman: thanks again for your support.
in this case, I decided to avoid your solution because, after the definition of this function handle i need to sum it with other function handles and integrate them. In fact, after calling a double integration and after separating the multiple conditions, this message appears:
Operands to the || and && operators must be convertible to logical scalar values.
any other suggestion to solve this other trouble?
Dyuman Joshi
Dyuman Joshi am 21 Nov. 2023
Bearbeitet: Dyuman Joshi am 21 Nov. 2023
A silly mistake from my side - It should be a single ampersand i.e. & .
If there is still an error after making the aforementioned change, please share your code and state the error you get.
Gianluca Angelini
Gianluca Angelini am 21 Nov. 2023
Don't worry. You couldn't have known about integrals.
By the way all the problems aforementioned are fixed. Thanks again!
Dyuman Joshi
Dyuman Joshi am 21 Nov. 2023
Bearbeitet: Dyuman Joshi am 21 Nov. 2023
The anonymous function is vectorized and so, for the given comparison, I should have noted that it should be a single ampersand. (I really should be going to sleep earlier :P)
Nonetheless, glad to know that your issue is solved. Happy to have helped!

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (0)

Kategorien

Mehr zu Data Type Identification finden Sie in Hilfe-Center und File Exchange

Produkte

Version

R2018b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by